Download Finite groups, representation theory and combinatorial structures

Finite Groups, Representation Theory and
Combinatorial Structures ∗
J Moori
School of Mathematical Sciences, North-West University (Mafikeng)
Mmabatho, 2375, South Africa †
May 19, 2015
Abstract
We introduce background material from Finite Groups and Representation Theory of Finite Groups (Linear and Permutation Representations).
We introduce the reader to combinatorial structures such as Designs and
Linear Codes and will discuss some of their properties, we also give few examples. We aim to introduce two new methods for constructing codes and
designs from finite groups (mostly simple finite groups). We outline some
of recent collaborative work by the author with J D Key, B Rorigues and T
Le.
Keywords: Group, representation, character, simple groups, maximal
subgroups, conjugacy classes, designs, codes.
1
Introduction
Error-correcting codes that have large automorphism groups are useful in applications as the group can help in determining the code’s properties, and can be
useful in decoding algorithms: see Huffman [15].
In a series of 3 lectures given at the NATO Advanced Study Institute ”Information Security and Related Combinatorics” held in Croatia [28], we discussed
two methods for constructing codes and designs for finite groups (mostly simple
finite groups). The first method dealt with construction of symmetric 1-designs
and binary codes obtained from from the action on the maximal subgroups, of a
finite group G. This method has been applied to several sporadic simple groups,
for example in [18], [22], [23], [31], [32], [33] and [34]. The second method introduces a technique from which a large number of non-symmetric 1-designs could
be constructed. Let G be a finite group, M be a maximal subgroup of G and
Cg = [g] = nX be the conjugacy class of G containing g. We construct 1−(v, k, λ)
designs D = (P, B), where P = nX and B = {(M ∩ nX)y |y ∈ G}. The parameters v, k, λ and further properties of D are determined. We also study codes
∗ AMS
Subject Classification (2000): 20D05, 05B05.
from CIMPA, NRF, University of North-Westl and AIMS are acknowledged.
† Supports
1
2
PERMUTATION GROUPS
2
associated with these designs. In Subsections 5.1, 5.2 and 5.3 we apply the second
method to the groups A7 , P SL2 (q) and J1 respectively.
Our notation will be standard, and it is as in [2] for designs and codes. For
groups we use ATLAS [5]. For the structure of finite simple groups and their
maximal subgroups we follow the ATLAS notation. The groups G.H, G : H,
and G· H denote a general extension, a split extension and a non-split extension
respectively. For a prime p, pn denotes the elementary abelian group of order pn .
If G is a group and M is a G-module, the socle of M , written Soc(M ), is the
largest semi-simple G-submodule of M . It is the direct sum of all the irreducible
G-submodules of M . Determination of Soc(V ) for each of the relevant full-space
G-modules V = F n is highly desirable.
2
2.1
Permutation Groups
Permutation Representations
Theorem 2.1 (Cayley) Every group G is isomorphic to a subgroup of SG . In
particular if |G| = n, then G is isomorphic to a subgroup of Sn .
Proof: For each x ∈ G, define Tx : G −→ G by Tx (g) = xg. Then Tx is one-to-one
and onto; so that Tx ∈ SG . Now if we define τ : G −→ SG by τ (x) = Tx , then τ
is a monomorphism. Hence G ∼
= Image(τ ) ≤ SG . Definition 2.1 The homomorphism τ defined in Theorem 2.1 is called the left
regular representation of G.
Note: Cayley’s Theorem is not that useful when the group G is large or when G
is simple. Following results (Theorem 2.3 and Corollary 2.4 ) provide substantial
improvement over Cayley’s Theorem. Notice that A5 ≤ S5 and Cayley’s Theorem
asserts that A5 is also a subgroup of S60 .
Corollary 2.2 Let GL(n, F) denote the general linear group over a field F. If
G is a finite group of order n, then G can be embedded in GL(n, F), that is G is
isomorphic to a subgroup of GL(n, F).
Proof: Let Tx be as in Cayley’s Theorem. Assume that G = {g1 , g2 , · · · , gn }.
Let Px = (aij ) denote the n × n matrix given by aij = 1F if Tx (gi ) = gj and aij =
0F , otherwise. Then Px is a permutation matrix, that is a matrix obtained
from the identity matrix by permuting its columns. Define ρ : G −→ GL(n, F) by
ρ(x) = Px , then it is not difficult to check that ρ is a monomorphism. Note: If Pn denotes the set of all n × n permutation matrices, then Pn is a group
under the multiplication of matrices and Pn ∼
= Sn .
Example 2.1 Consider the Klein four group V4 = {e, a, b, c, }. Then we have
Ta (e) = a.e = a, Ta (a) = a2 = e, Ta (b) = ab = c, Ta (c) = ac = b;
Tb (e) = b, Tb (a) = c, Tb (b) = e, Tb (c) = a;
Tc (e) = c, Tc (a) = b, Tc (c) = e, Tc (b) = a.
2
PERMUTATION GROUPS
3
Hence the permutation matrices are



0
0 1 0 0
 0
 1 0 0 0 


Pe = I4 , Pa = 
 0 0 0 1  , Pb =  1
0
0 0 1 0
0
0
0
1
1
0
0
0

0
1 
,
0 
0

0
 0
Pc = 
 0
1
0
0
1
0
0
1
0
0
So that V4 ∼
= {I4 , Pa , Pb , Pc } ≤ GL(4, F)
Exercise 2.1 (i) Show that for n ≥ 2, Sn is isomorphic to a subgroup of An+2 .
(ii) Use part (i) to show that A∞ contains an isomorphic copy of every finite
group. (See below for the definition of A∞ .)
Definition 2.2 Let X = N and let F be the set of all α ∈ SX such that α moves
finitely many elements of X. Then F ≤ SX . Let A∞ denote the subgroup of F
generated by all 3-cyles of F. It can be shown that A∞ is a simple group.
Theorem 2.3 (Generalized Cayley Theorem) Let H be a sugroup of G and
let X be the set of all left cosets of H in G. Then there is a homomorphism
ρ : G −→ SX such that
\
Ker(ρ) =
gHg −1 .
g∈G
Proof: For any x ∈ G, define ρx : X −→ X by ρx (gH) = x(gH). Then ρx is
well-defined, one-to-one and onto. So that ρx ∈ SX . Now define ρ : G −→ SX by
ρ(x)
for all x ∈ G. Then ρ is a homomorphism. We claim that Ker(ρ) =
T = ρx −1
.
g∈G gHg
Let x ∈ Ker(ρ). Then ρx = ρ(x) is the identity permutation on X. Hence
ρx (gH) = gH for all g ∈ G. So that xgH = gH, ∀g ∈ G. SoTg −1 xg ∈ H, ∀g ∈ G.
This implies that x ∈ gHg −1 , ∀g ∈ G. Thus Ker(ρ) ⊆ g∈G gHg −1 . Now if
T
x ∈ g∈G gHg −1 , then x ∈ gHg −1 , ∀g ∈ G. So that xgH = gH for all g ∈ G, that
T
is ρx is the identity permutation. Hence x ∈ Ker(ρ), so g∈G gHg −1 ⊆ Ker(ρ).
Definition 2.3 The homomorphism ρ defined above (Theorem 2.3) is called the
permutation
T representation of G on the left cosets of H in G. The kernel of
ρ, Ker(ρ) = g∈G gHg −1 , is called the core of H in G.
Exercise 2.2 If ρ is the permutation representation of G on the left cosets of H
in G, then show that
(i) Ker(ρ) ≤ H, (ii) G/Ker(ρ) is isomorphic to a subgroup of SX , where
X = G/H = {gH | g ∈ G}.
Corollary 2.4 If G is an infinite group such that contains a proper subgroup of
finite index, then G contains a proper normal subgroup of finite index.
Proof: Let H ≤ G such that [G : H] = n. Let X = G/H be the set of all left
cosets of H in G. Then
T |X| = n and there is a homomorphism ρ : G −→ Sn
such that Ker(ρ) = g∈G gHg −1 . Since G/Ker(ρ) is isomorphic to a subgroup
of Sn , G/Ker(ρ) is finite. Obviously Ker(ρ) G, and since Ker(ρ) ≤ H < G,
Ker(ρ) 6= G. Note that Ker(ρ) 6= {1G }. 
1
0 
.
0 
0
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4
Corollary 2.5 If G is a simple group containing a proper subgroup H of finite
index n, then G is isomorphic to a subgroup of Sn .
Proof: ByTTheorem 2.3, there exists a homomorphism ρ : G −→ Sn such that
Ker(ρ) = g∈G gHg −1 and Ker(ρ) ≤ H. Since Ker(ρ) G and G is simple,
Ker(ρ) = G or Ker(ρ) = {1G }. Since H < G and Ker(ρ) ≤ H, Ker(ρ) 6= G.
Thus Ker(ρ) = {1G }. Hence ρ is a monomorphism; so that G ∼
= Image(ρ) ≤ Sn .
Exercise 2.3 Prove that if λ and ρ are left and right regular representations of
G, then λ(a) commutes with ρ(b) for all a, b ∈ G.
Exercise 2.4 (i)∗ Let G be a group of order 2m k, where k is odd. Prove that if
G contains an element of order 2m , then the set of all elements of odd order in G
is a normal subgroup. (Hint: Consider G as permutations via Cayley’s Theorem,
and show that it contains an odd permutation).
(ii) Show that a finite simple group of even order must have order divisible by
4.
Exercise 2.5 (Poincare) If H and K are subgroups of G having finite index, then
H ∩ K has finite index. (Hint: [G : H ∩ K] ≤ [G : H][G : K].)
Exercise 2.6 Let G be a finite group and H ≤ G with [G : H] = p, where p is
the smallest prime divisor of |G|. Prove that H is normal in G.
Exercise 2.7 Prove that A6 has no subgroup of prime index.
Definition 2.4 (Conjugate subgroups) Let G be a group and H ≤ G we define H g by
H g : = gHg −1 = {ghg −1 | h ∈ H}.
Then H g is called the conjugate of H by g. It is routine to check that H g ≤
G, ∀g ∈ G.
Definition 2.5 (Normalizer) If H ≤ G, the normalizer of H in G, denoted
by NG (H), is defined by
NG (H) : = {g | g ∈ G, gHg −1 = H}.
If H G, then NG (H) = G.
Exercise 2.8 (i) Show that G ≥ NG (H) H.
are subgroups of G, then NG (H) ≥ K.
(ii) If H K, where H and K
Theorem 2.6 Let G be a group and H ≤ G. Let X = {gHg −1
T | g ∈ G}. Then
there exists a homomorphism φ : G −→ SX such that Ker(φ) = g∈G gNG (H)g −1 .
Proof: Define φg : X −→ X by φg (g 0 Hg 0−1 ) = g(g 0 Hg 0−1 )g −1 . Then φg is welldefined and φg ∈ SX . Now define φ : G −→ SX by φ(g) = φg . Then φ is a
homomorphism: ∀a, g ∈ G we have φ(ab) = φab and
φab (gHg −1 )
=
ab(gHg −1 )b−1 a−1 = a(bgHg −1 b−1 )a−1 = a(φb (gHg −1 ))a−1
=
φa (φb (gHg −1 )) = (φa ◦ φb )(gHg −1 ),
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PERMUTATION GROUPS
5
hence φab = φa ◦ φb on X and φ is a homomorphism.
If g ∈ Ker(φ), then φ(g) = φg is the identity permutation on X. So ∀g 0 ∈
G we have φg (g 0 Hg 0−1 ) = g 0 Hg 0−1 . Therefore g(g 0 Hg 0−1 )g −1 = g 0 Hg 0−1 ; so
g 0−1 gg 0 Hg 0−1 g −1 g 0 = H, that is g 0−1 gg 0 H(g 0−1 gg 0 )−1 = H. Hence g 0−1 gg 0 ∈
NG (H) and
g ∈ g 0 NG (H)g 0−1 , ∀g 0 ∈ G. This shows that
T we deduce that
−1
Ker(φ) ⊆ g∈G g.NG (H)g . (1)
T
If a ∈ g∈G gNG (H)g −1 , then a ∈ gNG (H)g −1 for all g ∈ G. Thus there is
g 0 ∈ NG (H) such that a = gg 0 g −1 . Now we have, for all g ∈ G,
φa (gHg −1 )
=
agHg −1 a−1 = gg 0 g −1 gHg −1 gg 0−1 g −1
=
gg 0 Hg 0−1 g −1 = gHg −1 ,
since g 0 ∈ NG (H).TThis shows that φa is the identity on X. Thus a ∈ Ker(φ) and
hence Ker(φ) ⊇ g∈G gNG (H)g −1 . (2) Now from (1) and (2) we obtain that
T
Ker(φ) = g∈G gNG (H)g −1 . Note: The homomorphism φ given in Theorem 2.6, is called the permutation
representation of G on the conjugates of H.
Exercise 2.9 Prove that a subgroup H of G is normal if and only if it has only
one conjugate in G.
Exercise 2.10 If H and K are conjugate subgroups of G, then H ∼
= K. Give an
example to show that the converse may be false.
Exercise 2.11 if λ and ρ are left and right regular representations of S3 , show
that λ(S3 ) and ρ(S3 ) are conjugate subgroups of S6 .
Exercise 2.12 Let G be a finite group with proper subgroup H. Prove that G
is not the set-theoretic union of all conjugates of H. Give an example in which
H is not normal and this union is a subgroup.
Exercise 2.13 (i) Assume H < K < G. Show that NK (H) = NG (H) ∩ K.
(ii) Prove that NG (xHX −1 ) = xNG (H)x−1 .
Exercise 2.14 If H and K are subgroups of G, show that NG (H ∩K) ≥ NG (H)∩
NG (K). Give an example in which the inclusion is proper.
Exercise 2.15 ∗ Let G be an infinite group containing an element x 6= 1G having
only finitely many conjugates. Prove that G is not simple.
3
Permutation Groups
Definition 3.1 Let G be a group and X be a set. We say that G acts on X if
there exists a homomorphism ρ : G −→ SX . Then ρ(g) ∈ SX for all g ∈ G. The
action of ρ(g) on X, that is ρ(g)(x), is denoted by xg for any x ∈ X. We say that
G is a permutation group on X.
3
PERMUTATION GROUPS
6
Example 3.1 (i) If G ≤ SX , then obviously G acts on X naturally.
(ii) By Cayley’s theorem any group G acts on itself and the action is given by
ag = ga, ∀a ∈ G, for g ∈ G.
(iii) If H ≤ G, then G acts on G/H, the set of all left cosets of H in G. The
action is given by: for g ∈ G, (aH)g = gaH, ∀a ∈ G.
(iv) If H ≤ G, then G acts on the set of all conjugates of H in G by
(aHa−1 )g = g(aHa−1 )g −1 = gaHa−1 g −1 .
Definition 3.2 (Orbits) Let G act on a set X and let x ∈ X. Then the orbit
of x under the action G is defined by
xG : = {xg | g ∈ G}.
Theorem 3.1 Let G act on a set X. The set of all orbits of G on X form a
partition of X.
Proof: Define the relation ∼ on X by x ∼ y if and only if x = y g for some g ∈ G.
Then ∼ is an equivalence relation on X (check) and [x] = {xg | g ∈ G} = xG .
Hence the set of all orbits of G on X partitions X. Example 3.2 (i) If G acts on itself by the left regular representation, then ∀g ∈ G
we have g G = {g h | h ∈ G} = {hg | h ∈ G} = Gg = G. Hence under the action of
G, we have only one orbit, namely G itself.
(ii) If G acts on G/H, the set of left cosets of H in G, then ∀aH ∈ G/H we
have
(aH)G = {(aH)g | g ∈ G} = {gaH | g ∈ G} = G/H.
In this case we have only one orbit, namely G/H.
(iii) In the case when G acts on itself by conjugation, that is for g ∈ G we
have ∀x ∈ G xg : = gxg −1 , then
xG = {xg | g ∈ G} = {gxg −1 | g ∈ G} = [x]
the conjugacy class of x in G. Note that |xG | = |[x]| = [G : CG (x)]. In this case
the number of orbits is equal to the number of conjugacy classes of G.
(iv) If G acts on the set of all its subgroups by conjugation, that is H g =
gHg −1 , ∀g ∈ G, ∀H ≤ G, then for a fixed H in G we have
H G = {H g | g ∈ G} = {gHg −1 | g ∈ G}
the set of all conjugates of H in G. Later we will prove that the number of
conjugates of H in G is equal to [G : NG (H)]. Hence |H G | = [G : NG (H)]. In
this case the number of orbits of G is equal to the number of conjugacy classes of
subgroups of G.
Definition 3.3 (Stabilizer) If G acts on a set X and x ∈ X then the stabilizer
of x in G, denoted by Gx is the set Gx = {g | xg = x}. That is Gx is the set of
elements of G that fixes x.
3
PERMUTATION GROUPS
7
Theorem 3.2 Let G act on a set X. Then
(i) Gx is a subgroup of G for each x ∈ X.
(ii) |xG | = [G : Gx ], that is the number of elements in the orbit of x is equal to
the index of Gx in G.
Proof: (i) Since x1G = x, 1G ∈ Gx . Hence Gx 6= Ø. Let g, h be two elements
−1
−1
of Gx . Then xg = xh = x. So (xg )h = (xh )h = x1G = x, and therefore
−1
xgh = x, ∀x ∈ X. Thus gh−1 ∈ Gx .
(ii) Since
xg = xh
−1
⇔ hg −1 ∈ Gx
⇔
x = xhg
⇔
(Gx )g = (Gx )h,
the map γ : xG −→ G/Gx given by γ(xg ) = (Gx )g is well-defined and one-to-one.
Obviously γ is onto. Hence there is a one-to-one correspondence between xG and
G/Gx . Thus |xG | = |G/Gx |. Exercise 3.1 Let G act on a set X. If y = xg for some x, y ∈ X, show that
g −1 Gx g = Gxg = Gy .
Corollary 3.3 If G is a finite group acting on a finite set X then ∀x ∈ X, |xG |
divides |G|.
Proof: By Theorem 3.2 we have |xG | = [G : Gx ] = |G|/|Gx |. Hence |G| =
|xG | × |Gx |. Thus |xG | divides |G|.
Theorem 3.4 (Applications of Theorem 3.2) (i) If G is a finite group, then
∀g ∈ G the number of conjugates of g in G is equal to [G:CG (g)].
(ii) If G is a finite group and H is a subgroup of G, then the number of
conjugates of H in G is equal to [G:NG (H)].
Proof: (i) Since G acts on itself by conjugation, using Theorem 3.2 we have
|g G | = [G:Gg ]. But since
g G = {g h | h ∈ G} = {hgh−1 | h ∈ G} = [g]
and
Gg = {h ∈ G | g h = g} = {h ∈ G | hgh−1 = g} = {h ∈ G | hg = gh} = CG (g),
we have
|g G | = |[g]| = [G:Gg ] = [G:CG (g)] =
|G|
.
|CG (g)|
(ii) Let G act on the set of all its subgroups by conjugation. Then by Theorem
3.2 we have |H G | = [G:GH ]. Since H G = {H g | g ∈ G} = {gHg −1 | g ∈ G} = [H]
and GH = {g ∈ G | H g = H} = {g ∈ G | gHg −1 = H} = NG (H) we have
|[H]| = |H G | = [G:GH ] = [G:NG (H)] = |NG|G|
(H)| . Theorem 3.5 (Cauchy - Frobenius ) Let G be a finite group acting on a finite
set X. Let n denote the number of orbits of G on
P X. Let F (g) denote the number
1
of elements of X fixed by g ∈ G. Then n = |G|
g∈G F (g).
3
PERMUTATION GROUPS
8
P
Proof: Consider S = g∈G F (g). Let x ∈ X. Since there are |Gx | elements in
G that fix x, x is counted |Gx | times in S. If ∆ = xG , then ∀y ∈ ∆ we have
|∆| = |xG | = |y G | = [G:Gx ] = [G:Gy ]. Hence |Gx | = |Gy |. Thus ∆ contributes
[G:Gx ].|Gx | to the sum S. But [G:Gx ].|Gx | = |G| is independent to the choice of
∆ and hence each orbit of G on X contributes |G| to the sum S. Since we have n
orbits, we have S = n|G|. Definition 3.4 (Transitive Groups) Let G be a group acting on a set X. If G
has only one orbit on X, then we say that G is transitive on X, otherwise we
say that G is intransitive on X. If G is transitive on X, then xG = X∀ x ∈ X.
This means that ∀x, y ∈ X, ∃g ∈ G such that xg = y.
Note: If G is a finite transitive group acting on a finite set X, then Theorem 3.2
(ii) implies that |xG | = |X| = |G|/|Gx |. Hence |G| = |X| × |Gx |.
Definition 3.5 (Multiply Transitive Groups) Let G act on a set X and let
|X| = n and 1 ≤ k ≤ n be a positive integer. We say that G is k - transitive on
X if for every two ordered k - tuples (x1 , x2 , · · · , xk ) and (y1 , y2 , · · · , yk ) with xi 6=
xj and yi 6= yj for i 6= j there exists g ∈ G such that xi g = yi for i = 1, 2, · · · , k.
The transitivity introduced in Definition 3.4 is the same as 1 - transitive.
Exercise 3.2 Let G be a group acting on a set X. Assume that |X| = n. Let
1 ≤ k ≤ n be a positive integer.
(i) Show that if G is k - transitive, then G is also (k − 1) transitive, when
k > 1.
(ii) If ∃H ≤ G such that H is k - transitive on X, then G is also k - transitive.
Exercise 3.3 Let G be a transitive group on a set X, |X| = k ≥ 2. Show that
G is k - transitive on X if and only if Gx is (k − 1) - transitive on X − {x}, for
every x ∈ X.
Theorem 3.6 If G is a k - transitive group on a set X with |X| = n, then
|G| = n(n − 1)(n − 2) · · · (n − k + 1)|G[x1 ,x2 ,··· ,xk ] |
for every choice of k- distinct x1 , x2 , · · · , xk ∈ X, where G[x1 ,x2 ,··· ,xk ] denote the
set of all elements g in G such that xi g = xi , 1 ≤ i ≤ k.
Proof: Let x1 ∈ X. Then since G is k - transitive, we have |G| = n × |Gx1 | (1)
and Gx1 is (k−1) - transitive, by Exercise 3.3, on X −{x1 }. Choose x2 ∈ X −{x1 }.
Then since Gx1 is (k − 1) - transitive on X − {x1 } we have |Gx1 | = |X − {x1 }| ×
|(Gx1 )x2 |, that is |Gx1 | = (n − 1) × |G[x1 ,x2 ] | and G[x1 ,x2 ] is (k − 2) - transitive on
X − {x1 , x2 }. (2) Notice that (1) and (2) imply that |G| = n(n − 1) × |G[x1 ,x2 ] |.
If we continue this way, we will get
|G| = n(n − 1)(n − 2) · · · (n − k + 1)|G[x1 ,x2 ,··· ,xk ] |.
Theorem 3.7 Let G act transitively on a finite set X with |X| > 1. Then there
exists g ∈ G such that g has no fixed points.
4
REPRESENTATION THEORY OF FINITE GROUPS
9
Proof: By the Cauchy - Frobenius theorem we have
1=n
=
1 X
F (g)
|G|
=
1
[F (1G ) +
|G|
g∈G
=
1
[|X| +
|G|
X
F (g)]
g∈G−{1G }
X
F (g)].
g∈G−{1G }
If F (g) > 0 for all g ∈ G, then we have
1=
1
[|X| +
|G|
X
g∈G−{1G }
F (g)] ≥
1
[|X| + |G| − 1]
|G|
≥ 1+
|X| − 1
> 1,
|G|
which is a contradiction. Hence ∃ g ∈ G such that F (g) = 0. Exercise 3.4 Let G be a group of permutations on a set X and let x, y ∈ X. If
xt = y for some t ∈ G, prove that Gx ∼
= Gy . (Hint: Use Exercise 3.1.)
Exercise 3.5 Use Cauchy - Frobenius to prove Lagrange’s Theorem. (Hint: Consider the left - regular action of G.)
Exercise 3.6 If G is a finite group
P and c is the number of conjugacy classes of
1
elements of G, show that c = |G|
x∈G |CG (x)|. (Hint: Consider the conjugation
action of G on its elements and use Cauchy - Frobenius Theorem).
Exercise 3.7 Let G be a finite group of order pn , where p is a prime. Assume
that G acts on a set X with p not dividing |X|. Prove that there exists x ∈ X
such that xg = x for all g ∈ G. [Hint: use Corollary 3.3.]
Exercise 3.8 Assume that V ia vector space of dimension n over Zp and GL(n, p)
is the corresponding general linear group acting on V . If G is a subgroup of
GL(n, p) with |G| = pm , prove that there exists a non-zero vector v ∈ V such
that gv = v for all g ∈ G. [Hint: since G ≤ GL(n, p), G acts on V .]
4
4.1
Representation Theory of Finite Groups
Basic Concepts
Definition 4.1 Let G be a group. Let f : G −→ GL(n, F) be a homomorphism.
Then we say that f is a Matrix Representation of G of degree n (or dimension
n), over the field F.
If Ker(f ) = {1G }, then we say that f is a faithful representation of G. In this
situation G ∼
= Image(f ); so that G is isomorphic to a subgroup of GL(n, F).
4
REPRESENTATION THEORY OF FINITE GROUPS
10
Example 4.1 (i) The map f : G −→ GL(1, F) given by f (g) = 1F for all g ∈ G
is called the trivial representation of G over F. Notice that GL(1, F) = F∗ .
(ii) Let G be a permutation group acting on a finite set X, where X =
{x1 , x2 , · · · , xn }. Define π : G −→ GL(n, F) by π(g) = πg for all g ∈ G, where πg
is the permutation matrix induced by g on X. That is πg = (aij ) an n × n
matrix having 0F and 1F as entries in such way that
aij
=
1F
if g(xi ) = xj
=
0F otherwise.
Then π is a representation of G over F, and π is called the permutation representation of G on X.
(iii) Take X = G in part (ii). Define a permutation action on G by g : x −→ xg
for all x ∈ G. Then the associated representation π is called the right regular
representation of G.
Exercise 4.1 Let N G. Assume that ρ is a representation of G/N . Define
ρ : G −→ GL(n, F), where n is the degree of ρ̂, by ρ(g) = ρ(gN ). Then show that
ρ is a representation of G.
Exercise 4.2 Let N G. Assume that ρ is a representation of degree n on G.
If N ≤ Ker(ρ), then show that the mapping ρ : G/N −→ GL(n, F) given by
ρ(gN ) = ρ(g) is a representation of G/N.
Theorem 4.1 Let G be a group. Then the derived subgroup G0 lies in the kernel
of any representation of G of degree 1.
Proof: Assume that f : G −→ GL(1, F) is a representation of degree one of G.
Let a, b ∈ G. Then
f (aba−1 b−1 ) = f (a)f (b)[f (a)]−1 [f (b)]−1 .
Since GL(1, F) = F ∗ is abelian we have
f (aba−1 b−1 ) = f (a)[f (a)]−1 f (b)[f (b)]−1 = 1F .
Thus [a, b] = aba−1 b−1 ∈ Kerf. Since G0 is generated by the set of all commutators, G0 ⊆ Ker(f ). Exercise 4.3 Let F = GF (q) be the Galois Field of q elements, where q = pk
for some prime p. Show that |GL(n, F)| = (q n − 1).(q n − q) · · · (q n − q n−1 ).
Definition 4.2 (Special Linear Group SL(n, F)) Let F be any field.
SL(n, F) = {A | A ∈ GL(n, F), det(A) = 1F }.
Then it is not difficult to show that SL(n, F) ≤ GL(n, F).
Theorem 4.2 Let F = GF (q) with q = pk for some prime p. Then
SL(n, F) GL(n, F)
and
|SL(n, F)| = |GL(n, F)|/(q − 1).
4
REPRESENTATION THEORY OF FINITE GROUPS
11
Proof: Let ρ : GL(n, F) −→ F ∗ be given by ρ(A) = det(A), for all A ∈ GL(n, F).
Then ρ is a homomorphism (check ). If a ∈ F∗ , then


a 0 0 ··· 0
 0 1 0 ··· 0 




ρ :  0 0 1 · · · 0  7−→ a,
 ..

 .

0
0
0
···
1
so that ρ is onto. We also have Ker(ρ) = {A | A ∈ GL(n, F), det(A) = 1F } =
SL(n, F). Since Ker(ρ)GL(n, F), SL(n, F)GL(n, F). Now since GL(n, F)/Ker(ρ) ∼
=
Image(ρ), we have that GL(n, F)/SL(n, F) ∼
= F ∗ . Hence
|GL(n, F)/SL(n, F)| = |F ∗ | = q − 1.
Corollary 4.3 If ρ : G −→ GL(n, F) is a representation of G, then ρ(g) ∈
SL(n, F) for all g ∈ G0 .
Proof: Let h = [a, b] be a commutator in G. Then we have ρ(h) = ρ(aba−1 b−1 ) =
ρ(a)ρ(b)ρ(a−1 )ρ(b−1 ). Now since
det(ρ(h))
= det(ρ(a)).det(ρ(b)).det(ρ(a−1 ))det(ρ(b−1 ))
=
det(ρ(aa−1 )).det(ρ(bb−1 ))
=
det(ρ(1G )).det(ρ(1G ))
=
det(In ).det(In ) = 1,
we have ρ(h) ∈ SL(n, F). Exercise 4.4 (Special triangular Group) Let ST L(N, F) denote the set of
all invertible lower triangular n × n matrices whose diagonal entries are all 1F .
Then ST L(n, F) ≤ GL(n, F). Show that if F = GF (q) where q = pk for some
prime p, then ST L(n, F) is Sylow p-subgroup of GL(n, F).
Definition 4.3 (Characters) Let f : G −→ GL(n, F) be a representation of G
over the field F. The function χ : G −→ F defined by χ(g) = tr(f (g)) is called the
character of f.
Definition 4.4 (Class functions) If φ : G −→ F is a function that is constant
on conjugacy classes of G, that is φ(g) = φ(xgx−1 ), ∀x ∈ G, then we say that φ
is a class function.
Lemma 4.4 A character is a class function.
Proof: Let χ be a character of G. Then χ is afforded by a representation ρ : G −→
GL(n, F). Let g ∈ G; then ∀x ∈ G we have
χ(xgx−1 )
=
tr(ρ(xgx−1 ))
=
tr(ρ(x).ρ(g).ρ(x−1 ))
=
=
tr(ρ(x).ρ(g).[ρ(x)]−1 )
tr(ρ(g)), see Note 5.1.1 below
=
χ(g).
4
REPRESENTATION THEORY OF FINITE GROUPS
12
Note: Similar matrices have the same trace. If A = (aij ) and B = (bij ) are two
matrices, then
tr(AB) =
n X
n
n X
n
X
X
(
aij bji ) =
(
bji aij ) = tr(BA).
i=1 j=1
j=1 i=1
Now if B = P AP −1 , then tr(B) = tr(P AP −1 ) = tr(P −1 P A) = tr(A). Note: If
χ is a character afforded by a representation ρ : G −→ GL(n, F), then χ is not
linear (in general) :
χ(gg 0 ) = tr(ρ(gg 0 )) = tr(ρ(g)ρ(g 0 )) 6= tr(ρ(g)) × tr(ρ(g 0 )) = χ(g) × χ(g 0 ).
Later we will show that χ is linear if and only if deg(ρ) = 1.
Definition 4.5 (Equivalent Representations) Two representations
ρ, φ : G −→ GL(n, F) are said to be equivalent if there exists a n × n matrix P
over F such that P −1 ρ(g)P = φ(g), ∀g ∈ G.
Since similar matrices have the same trace, it follows that equivalent representations have the same character.
Theorem 4.5 Equivalent representations have the same character.
Proof: Let χ1 and χ2 be characters afforded by ρ1 and ρ2 two representations of
degree n over a field F. Assume that ρ1 is equivalent to ρ2 . Then there is a n × n
matrix P such that P −1 ρ1 (g)P = ρ2 (g), ∀g ∈ G. Now ∀g ∈ G we have
χ2 (g) = tr(ρ2 (g)) = tr(P −1 ρ1 (g)P ) = tr(ρ1 (g)) = χ1 (g).
Hence χ1 = χ2 . Definition 4.6 Let S be a set of (n × n) matrices over F. We say that S is
reducible if ∃ m, k ∈ N, and there exists P ∈ GL(n, F) such that ∀A ∈ S we
have
B 0
−1
P AP =
C D
where B is an m × m matrix, D and C are k × k and k × m matrices respectively.
Here 0 denotes the m × k zero matrix.
If there is no such P , we say that S is irreducible.
If C = 0, the zero k × m matrix, for all A ∈ S, then we say that S is fully
reducible.
We say that S is completely reducible if ∃ P ∈ GL(n, F) such that


B1 0
· · 0
 0
B2 0 · 0 
 , ∀A ∈ S,
P AP −1 = 
 ·
·
· · · 
0
0
· · Bk
where each Bi is irreducible.
4
REPRESENTATION THEORY OF FINITE GROUPS
13
Example 4.2 Let F = C; consider
a −b
S=
| a, b ∈ C .
b
a
1 −i
1 i
−1
Then S is a reducible set over C. Let P =
. Then P =
0
1
0 1
and
a −b
a + ib
0
P −1
P =
, ∀a, b ∈ C.
b
a
b a − ib
1 i
In fact we can show that S is fully reducible. For this let P =
. Then
i1
1 −i
a −b
a − ib
0
and P −1
P =
.
P −1 = 21
−i
1
b
a
0 a + ib
1 1
Exercise 4.5 Let S =
. Show that S is reducible, but it is not fully
0 1
reducible.
Definition 4.7 Let f : G −→ GL(n, F) be a representation of G over F. Let
S = Im(f ) = {f (g) | g ∈ G} . Then S ⊆ GL(n, F). We say that f is reducible,
fully reducible, or completely reducible if S is reducible, fully reducible or
completely reducible.
Definition 4.8 (Sum of representations) let ρ : G −→ GL(n, F) and φ : G −→
GL(m, F) be two representations of G over F. Define
ρ + φ : G −→ GL(n + m, F) by
ρ(g) 0n×m
(ρ + φ)(g) :=
= ρ(g) ⊕ φ(g),
0m×n
φ(g)
for all g ∈ G. Then ρ + φ is a representation of G over F, of degree n + m.
If χ1 and χ2 are the characters of ρ and φ respectively, and if χ is the character
of ρ + φ, then ∀g ∈ G we have
ρ(g)
0
χ(g) = trace
= tr(ρ(g)) + tr(φ(g)) = χ1 (g) + χ2 (g) = (χ1 + χ2 )(g).
0 φ(g)
Hence χ = χ1 + χ2 .
Example 4.3 Let G =< a, b >such thata2 = b2 =1G and ab
= ba. Define
1
0
−1 0
f : G −→ GL(2, C) by f (a) =
, f (b) =
. Then f is a
0 −1
0 1
faithful representation of degree 2. It is not difficult to see that f is completely
reducible
Exercise 4.6 Represent the permutations of S3 as permutation matrices. Calculate the character of this representation.
4
REPRESENTATION THEORY OF FINITE GROUPS
14
Theorem 4.6 (Maschke’s theorem) Let G be a finite group. Let f be a representation of G over a filed F whose characteristic is either 0 or is a prime that
does not divide |G|. If f is reducible, then f is fully reducible.
Proof: In fact we will show that if there is a matrix P such that ∀g ∈ G
A(g)
0
−1
P f (g)P =
B(g) C(g)
then there is a matrix Q such that
∀g ∈ G, Q−1 f (g)Q =
A(g)
0
0 C(g)
.
Ir 0
where r and s are the degrees of
T Is
A(g) and C(g) respectively, and T is an s × r matrix. We need to determine T
such that L is an invertible matrix over F independent of g and
A(g)
0
, ∀g ∈ G. (1)
L−1 f (g)L =
0 C(g)
Let f (g) = P −1 f (g)P and let L =
Then Q = P L. Relation (1) implies that
A(g)
0
Ir 0
Ir
=
B(g) C(g)
T Is
T
that is
A(g)
B(g) + C(g)T
0
C(g)
0
Is
=
A(g)
0
0 C(g)
A(g)
0
T.A(g) C(g)
,
.
Hence we find that
B(g) + C(g)T = T.A(g), ∀g ∈ G.
(2).
Since f is a matrix representation of G, f (gh) = f (g).f (h) for all g, h ∈ G. So
for all g and h in G we have
A(gh)
0
A(g)
0
A(h)
0
=
,
B(gh) C(gh)
B(g) C(g)
B(h) C(h)
that is
A(gh)
0
B(gh) C(gh)
=
A(g)A(h)
0
B(g)A(h) + C(g)B(h) C(g)C(h)
We obtain the relations:
(i) A(gh) = A(g)A(h)
(ii) C(gh) = C(g)C(h)
(iii) B(gh) = B(g)A(h) + C(g)B(h).
.
4
REPRESENTATION THEORY OF FINITE GROUPS
15
Relations (i) and (ii) show that A and C are matrix representations of G over
F. By multiplying (iii) and A(h−1 ) we obtain that
B(gh)A(h−1 ) = B(g) + C(g)B(h)A(h−1 ) ∀g, h ∈ G.
(3)
If we fix g and let h runs over all elements of G, then using (3) we get
X
X
X
B(gh)A(h−1 ) =
B(g) +
C(g)B(h)A(h−1 )
h∈G
h∈G
= |G|B(g) + C(g)
h∈G
X
B(h)A(h−1 ).
(4)
h∈G
Now let x = gh. Since h runs over all elements of G, so also does x. Hence
X
X
X
B(gh)A(h−1 ) =
B(x)A(x−1 g) =
B(x)A(x−1 )A(g)
x∈G
h∈G
=(
X
x∈G
B(x)A(x−1 ))A(g).
(5)
x∈G
Now relations (4) and (5) give
X
X
(
B(x)A(x−1 ))A(g) = |G|B(g) + C(g)(
B(h)A(h−1 )).
x∈G
(6)
h∈G
Since the characteristic of F does not divide |G|, |G| 6= 0F in F. Hence we can
divide both sides of relation (6) by |G|. We get
(
1 X
1 X
B(h)A(h−1 )).
B(x)A(x−1 ))A(g) = B(g) + C(g)(
|G|
|G|
x∈G
(7)
h∈G
Finally by comparing relations (7) and (2), if we let
T =
1 X
(
B(x)A(x−1 )),
|G|
x∈G
then T satisfies the relation (2). Theorem 4.7 [The general form of Maschke’s theorem] Let G be a finite group
and F a field whose characteristic is either 0 or is a prime that does not divide
|G|. Then every representation of G over F is completely reducible.
Proof: Let f be a representation of G over F. If f is irreducible, then it is
completely reducible. Hence assume that f is reducible. Then by Maschke’s
theorem f is fully reducible, and therefore for all g ∈ G, f (g) is similar to
A(g)
0
0 C(g)
.
4
REPRESENTATION THEORY OF FINITE GROUPS
16
Since A and B are representations of G over F , we can apply Maschke’s
theorem to these representations. Repeating this process we obtain that f (g) is
similar to


B1 (g) 0
· · 0
 0

B2 (g) 0 · 0

,
 ·

·
· · ·
0
0
· · Bk (g)
where Bi , 1 ≤ i ≤ k are all irreducible representations of G over F. Theorem 4.8 [Schur’s Lemma] Let ρ and φ be two irreducible representations,
of degree n and m respectively, of a group G over a field F. Assume that there
exists an m × n matrix P such that P ρ(g) = φ(g)P for all g ∈ G. Then either
P = 0m×n or P is non-singular so that ρ(g) = P −1 φ(g)P (that is ρ and φ are
equivalent representations of G).
Proof: Let r = rank(P ). Then there are non-singular matrices L and M such
that P = LEr M , where
0m−r×r 0m−r×n−r
Er =
,
Ir
0r×n−r
m×n
and L and M are m × m and n × n matrices respectively. Since for all g ∈ G we
have P ρ(g) = φ(g)P, we obtain that
LEr M ρ(g) = φ(g)LEr M.
Hence
Er M ρ(g)M −1 = L−1 φ(g)LEr .
(1)
Using the relation (1), we can partition the matrices M ρ(g)M −1 and L−1 φ(g)L
in the following way, provided that r 6= 0, and m 6= r or n 6= r:
A(g) B(g)
−1
,
M ρ(g)M =
C(g) D(g) n×n
and
L−1 φ(g)L =
A0 (g) B 0 (g)
C 0 (g) D0 (g)
,
m×m
where A(g) is r × r, B(g) is r × n − r, C(g) is n − r × r, D(g) is n − r × n − r,
A0 (g) is m − r × m − r, B 0 (g) is m − r × r, C 0 (g) is r × m − r, D0 (g) is r × r. We
can easily deduce that
0m−r×r 0m−r×n−r
Er M ρ(g)M −1 =
A(g)
B(g)
and
L−1 φ(g)LEr =
B 0 (g) 0m−r×n−r
D0 (g) 0r×n−r
Now using the relation (1) we must have
0
0m−r×r 0m−r×n−r
B (g)
=
A(g)
B(g)
D0 (g)
.
0m−r×n−r
0r×n−r
.
4
REPRESENTATION THEORY OF FINITE GROUPS
17
Hence B 0 (g) = 0m−r×r and B(g) = 0r×m−r for all g ∈ G. This shows that ρ and
φ are reducible, which is a contradiction. Thus either r = 0 or m = n = r. If
r = 0, then P = 0m×n . If m = n = r, then P is invertible and ρ(g) = P −1 φ(g)P.
Definition 4.9 (Algebraically Closed Fields) . A field F is said to be Algebraically closed if every polynomial equation p(x) = 0F with P (x) ∈ F [x] has
all its roots in F.
For example the complex field C is an algebraically closed field by the Fundamental Theorem of Algebra. The first proof for the fundamental theorem of
algebra was given by Gauss in his doctoral dissertation in 1799 at the age of 22
(in fact gauss gave several independent proofs, he published his last proof in 1849
at the age of 72).
For an algebraically closed field, the Schur’s Lemma (Theorem 4.8) has the
following noteworthy corollary.
Corollary 4.9 If ρ is an irreducible representation of degree n of a group G over
an algebraically closed field F , then the only matrices which commute with all
matrices ρ(g), g ∈ G, are the scalar matrices aIn , a ∈ F.
Proof: Let P be an n × n matrix such that P ρ(g) = ρ(g)P , for all g ∈ G. Then
for any a ∈ F we have
(aIn − P )ρ(g) = ρ(g)(aIn − P ), f or all g ∈ G
(1)
Let m(x) = det(xIn − P ) be the characteristic polynomial of P. Since m(x) is
a polynomial over F and F is algebraically closed, there is a0 ∈ F such that
m(a0 ) = 0F . Hence det(a0 In − P ) = 0. So that a0 In − P is singular. Now using
relation (1) and Schur’s Lemma, we must have a0 In − P = 0. Thus P = a0 In . Exercise 4.7 Let G be a finite group and ρ and φ be representations of degrees n
and m respectively, over a a field F. Assume that char(F ) does not divide the order
of G. Let S be an m × n matrix over F . Show that Sρ(g) = φ(g)S
P for all g in G if
and only if there is an m × n matrix T over F such that S = g∈G φ(g −1 )T ρ(g).
Exercise 4.8 Maschke’s Theorem becomes false if the hypothesis that char(F )
does not divide the order of G is omitted. Let F = GF (2) and G =< a > be
acyclic
group
of order 2. Define ρ : G → GL(2, F ) by ρ(1G ) = I2 and ρ(a) =
1 1
. Show that ρ is reducible but not fully reducible. (Hint use Exercise
0 1
5.1.5.)
Exercise 4.9 Show that Corollary 5.1.9 is false if F = R.
Theorem 4.10 Let ρ and φ be two inequivalent irreducible representations, of
degrees n and m respectively, of a group G over a field F . If T is an m × n matrix
over F , then
X
φ(g −1 )T ρ(g) = 0m×n .
g∈G
5
CHARACTERS OF FINITE GROUPS
18
φ(g −1 )T ρ(g). Then for any x ∈ G we have
X
X
Sρ(x) =
φ(g −1 )T ρ(gx) =
φ(x)φ(x−1 g −1 )T ρ(gx) =
Proof: Let S =
P
g∈G
g∈G
φ(x)(
X
g∈G
φ((gx)−1 )T ρ(gx)) = φ(x)(
g∈G
X
φ(z −1 )T ρ(z)) = φ(x)S.
z∈G
Since ρ and φ are inequivalent irreducible representations of G, by Schur’s
lemma we must have S = 0m×n .
Definition 4.10 Let G be a finite group and F a field such that char(F ) does
not divide the order of G. If ρ and φ are two functions from G into F , we define
an inner product <, > by the following rule:
1 X
ρ(g)φ(g −1 ),
< ρ, φ >=
|G|
g∈G
where
1
|G|
stands for |G|−1 in F.
Theorem 4.11 The inner product <, > defined above is bilinear and symmetric:
(i) < ρ1 + ρ2 , φ >=< ρ1 , φ > + < ρ2 , φ >,
(ii) < ρ, φ1 + φ2 >=< ρ, φ1 > + < ρ, φ2 >,
(iii) < aρ, φ >= a < ρ, φ >=< ρ, aφ >, for all a ∈ F,
(iv) < ρ, φ >=< φ, ρ > .
Proof: the bilinear properties (i), (ii) and (iii) are easy to verify. Let us prove
the the symmetry:
1 X
1 X
1 X
ρ(g)φ(g −1 ) =
ρ(g −1 )φ(g) =
φ(g)ρ(g −1 ) =< φ, ρ > .
< ρ, φ >=
|G|
|G|
|G|
g∈G
g∈G
g∈G
Note: If ρ : G −→ F ∗ is a group homomorphism, then
1 X
1 X
1 X
1
< ρ, ρ >=
ρ(g)ρ(g −1 ) =
ρ(1G ) =
1F =
×(|G|1F ) = 1F .
|G|
|G|
|G|
|G|
g∈G
5
g∈G
g∈G
Characters of Finite Groups
In this section, unless explicit exception is made, the group G will be finite and
all representations and matrices will be over the complex field C.
Note: By the general form of Maschke’s theorem (Theorem 4.7), all representations of G are completely reducible.
Note: If If ρ : G −→ GL(n, C) is a
representation of G, then we denote the (i, j) entry of ρ(g) by ρij (g). Hence we
can regard ρij is a map from G into C.
5
CHARACTERS OF FINITE GROUPS
19
Theorem 5.1 [Orthogonality of irreducible representations] Let G be a finite
group and ρ and φ two irreducible representations of G.
(i) If ρ and φ are inequivalent, then < ρrs , φij >= 0, for all i, j, r, s.
(ii) < ρrs , ρij >= δis δjr /deg(ρ).
Proof: (i) Using Theorem 5.1.10 we have
X
φ(g −1 )Ejr ρ(g) = 0m×n ,
(1)
g∈G
where Ejr is the m × n matrix with (j, r) entry 1 and other entries 0, with
n = deg(ρ) and m = deg(φ). Now from (1) we get
1 X
φ(g −1 )Ejr ρ(g) = 0m×n .
|G|
(2)
g∈G
Since the (i, s) entry of the left hand-side of the relation (2) is
1 X
φij (g −1 )ρrs (g) =< φij , ρrs >,
|G|
g∈G
we have < φij , ρrs >= P
0.
1
−1
)Ejr ρ(g). Then for any x ∈ G we have (see
(ii) Let Sjr = |G|
g∈G ρ(g
Exercise 5.1.7) Sjr ρ(x) = ρ(x)Sjr , and from Corollary 5.1.9 we deduce that Sjr
is an scalar matrix. Hence let Sjr = λjr In , where λjr ∈ C. Then we have
λjr In =
1 X
ρ(g −1 )Ejr ρ(g).
|G|
(3)
g∈G
By comparing the (i, s) entry of the left and right hand-side of (3), we get
λjr δis =
1 X
ρij (g −1 )ρrs (g).
|G|
g∈G
That is
λjr δis =< ρij , ρrs > .
Since < ρij , ρrs >=< ρrs , ρij >, we get
λjr δis =< ρij , ρrs >=< ρrs , ρij >= λsi δrj .
(4)
Now if i 6= s or j 6= r, we have < ρij , ρrs >= 0 and (ii) holds. Suppose that
i = s and j = r. Then by (4) we have
< ρij , ρji >= λjj = λii .
(5)
Hence we have
λ11 = λ22 = · · · = λnn = λ ∈ C,
5
CHARACTERS OF FINITE GROUPS
so that
nλ =
n
X
λii =
i=1
=
n
X
20
< ρi1 , ρ1i >, by (5),
i=1
n
X
1 X
ρ1i (g −1 )ρi1 (g))
(
|G|
i=1
g∈G
=
n
X
1 X
(
ρ1i (g −1 )ρi1 (g)).
|G|
i=1
(6)
g∈G
Pn
Since i=1 ρ1i (g −1 )ρi1 (g) is the (1, 1)−entry of ρ(g −1 )ρ(g) and since ρ(g −1 )ρ(g) =
ρ(g −1 g) = ρ(1G ) = In , we have
n
X
ρ1i (g −1 )ρi1 (g) = 1.
i=1
Now relation (6) implies that
nλ =
1
1 X
1=
× |G| = 1.
|G|
|G|
g∈G
Hence λ =
1
n.
Therefore by (5) we get
< ρij , ρji >=
1
= δii δjj /deg(ρ).
n
Theorem 5.2 [Orthogonality of irreducible characters] Let G be a finite group
and ρ and φ two irreducible representations of G. If χρ and χφ are characters of
ρ and φ respectively, then
(i) < χρ , χφ >= 1 if ρ and φ are equivalent, and
< χρ , χφ >= 0 otherwise,
(ii) < χρ , χρ >= 1.
Proof: (i) Let n = deg(ρ) and m = deg(φ). Then we have
< χρ , χφ > =
1 X
χρ (g)χφ (g −1 )
|G|
g∈G
=
n
n
X
1 X X
{[
ρii (g)][
φjj (g −1 )]}
|G|
i=1
i=1
=
n X
m
X
1 X
[
ρii (g)φjj (g −1 )]
|G|
i=1 j=1
g∈G
g∈G
=
n X
m
X
i=1 j=1
< ρii , φjj > .
(1)
5
CHARACTERS OF FINITE GROUPS
21
If ρ and φ are inequivalent, then < ρii , φjj >= 0 by part (ii) of Theorem 5.1,
Hence using the relation (1) above we get < χρ , χφ >= 0.
If If ρ and φ are equivalent, then χρ = χφ by Theorem 5.1.5. Now we have
< χρ , χφ > =
=
< χρ , χρ >
n X
n
X
< ρii , ρjj >, by (1),
i=1 j=1
=
n
X
< ρii , ρii >=
i=1
=
n×
n
X
1
, by T heorem 5.2.1,
n
i=1
1
= 1.
n
(ii) As above. Corollary 5.3 Two irreducible representations of a finite group G are equivalent
if and only if they have the same characters.
Proof: Let ρ and φ be two irreducible representations of G. If ρ and φ are
equivalent then χρ = χφ by Theorem 5.1.5. Conversely assume that χρ = χφ .
Then by Theorem 5.2.2 (part (ii)) we have < χρ , χφ >= 1. Thus ρ and φ are
equivalent by Theorem 5.2.2. Note: Maschke’s theorem (Theorem 5.1.7) implies that if ρ is a representation
Pk
of G, then ρ is equivalent to i=1 ρi where ρi , s are irreducible representations of
Pk
G. We also have χρ = i=1 χρi .
Pk
Exercise 5.1 If ρ is a representation of G such that ρ is equivalent to i=1 ρi
where ρi , s are irreducible representations of G, then ρi are unique up to equivalence.
Theorem 5.4 (Generalisation of Corollary 5.2.3) Two representations of a
finite group G are equivalent if and only if they have the same characters.
Proof: Let ρ and φ be two representations of G. If ρ and φ are equivalent then
χρ = χφ by Theorem 5.1.5.
Conversely assume that χρ = χφ . Assume that an irreducible representation
ψi appears mi times inPρ and ni times in P
φ. Then adding dummy
Pk terms if
k
k
necessary, we have ρ ∼ i=1 mi ψi and φ ∼ i=1 ni ψi . Then χρ = i=1 mi χψi
Pk
Pk
Pk
and χφ = i=1 ni χψi . Since χρ = χφ , we have i=1 mi χψi = i=1 ni χψi . Hence
for any j we have
mj
=
< mj χψj , χψj >=<
k
X
mi χψi , χψj >
i=1
=
<
k
X
ni χψi , χψj >= nj .
i=1
Thus
Pk
i=1
mi ψi =
Pk
i=1
ni ψi . So ρ ∼ φ.
5
CHARACTERS OF FINITE GROUPS
22
Definition 5.1 (Irreducible Characters) If χρ is a character afforded by a
representation ρ of G, then we say that χρ is an irreducible character if ρ is
an irreducible representation. Notice that if χ is an irreducible character of G
and if φ is a representation of G such that χφ = χ, then φ is also irreducible by
Theorem 5.2.4.
Theorem 5.5 The set of all irreducible characters of G is a linearly independent
set over C.
Proof: Let χ1 , χ2 , · · ·, χm be a finite set of distinct irreducible characters of a
finite group G. assume that there are λ1 , λ2 , · · ·, λm in C such that
λ1 χ1 + λ2 χ2 + · · · + λm χm = 0,
(1)
the zero function from G into C. Since χi 6= χj for i 6= j, χi and χj are afforded
by inequivalent irreducible representations of G (by Theorem 5.2.4). Hence we
have < χi , χj >= 1 if i = j and 0 otherwise. Now using the relation (1) above we
get
0
= < 0, χj >=<
m
X
λi χi , χj >
i=1
= <
m
X
λi < χi , χj >= λj , 1 ≤ j ≤ m.
i=1
Therefore {χ1 , χ2 , · · ·, χm } is a linearly independent set over C. Theorem 5.6 If G has r distinct conjugacy classes of elements, then G has at
most r irreducible characters.
Proof: Let S = {[g1 ], [g2 ], · · ·, [gr ]} be the set of all conjugacy classes of elements
of G and let V be the vector space of functions from S into C. Define fi : S −→ C,
for 1 ≤ i ≤ r, by fi ([gi ]) = 1 and fi ([gj ]) = 0 if i 6= j. Then {f1 , f2 , · · ·, fr } is a
basis for V . Thus dim(V ) = r.
Let Irr(G) denote the set of all distinct irreducible characters of G. Since
a character is a class function, we can regard Irr(G) as a subset of V. Now
since Irr(G) is a linearly independent subset of V by Theorem 5.2.5, we have
|Irr(G)| ≤ dim(V ), that is |Irr(G)| ≤ r. Pk
Exercise 5.2 (i) If χ = i=1 λi χi , where χi are distinct irreducible characters
Pk
of G and λi are non-negative integers, show that < χ, χ >= i=1 λ2i .
(ii) If χ is a character of G, then show that χ is irreducible if and only if
< χ, χ >= 1.
Assume that {C1 , C2 , · · ·, Cr } is the set of all conjugacy classes of elements of
G with C1 = 1G . Unless otherwise stated gi will denote an arbitrary element of
the conjugacy class Ci and we put
hi = |Ci | = |G|/|CG (gi )|.
5
CHARACTERS OF FINITE GROUPS
23
Let Irr(G) = {χ1 , χ2 , · · ·, χk } be the set of all distinct irreducible characters of G with the assumption that χ1 is the character afforded by the trivial
representation ρ(g) = 1 for all g in G.
Theorem 5.7 We have the following
P
(i)
g∈G χ1 (g) = |G|,
P
(ii)
g∈G χi (g) = 0, if i 6= 1,
Pr
(iii)
j=1 hj χi (gj ) = δ1i |G|.
Proof: (i) Since χ1 (g) = 1 for all g in G, we have
X
X
χ1 (g) =
1 = |G|.
g∈G
g∈G
(ii) If i 6= 1, then < χi , χ1 >= 0. hence
1 X
1 X
χi (g)χ1 (g −1 ) =
χi (g).
0=
|G|
|G|
g∈G
g∈G
P
Thus g∈G χi (g) = 0 × |G| = 0.
P
(iii) If i = 1, then by part (i) we have g∈G χ1 (g) = |G| and hence
|G| =
X
χ1 (g) =
0 = δ1i =
hj χ1 (gj ).
j=1
g∈G
If i 6= 1, then by part (ii) we have
r
X
P
g∈G
X
χi (g) = 0 = δ1i and hence
χi (g) =
r
X
hj χi (gj ).
j=1
g∈G
Exercise 5.3 (i) Let ρ be an irreducible representation of G. Show that deg(ρ) =
1 if and only if ker(ρ) ≥ G0 .
(ii) Show that all irreducible representations of an abelian group are of degree
one.
Note: If χρ and χφ are two characters of G, we know that
1 X
< χρ , χφ >=
χρ (g)χφ (g −1 ).
|G|
g∈G
Hence
r
< χρ , χφ > =
=
=
1 X
hi χρ (gi )χφ (gi−1 )
|G| i=1
r
X
hi
χρ (gi )χφ (gi−1 )
|G|
i=1
r
X
i=1
1
χρ (gi )χφ (gi−1 ).
|CG (gi )|
5
CHARACTERS OF FINITE GROUPS
24
Example 5.1 Consider the symmetric group S3 . Representing the elements of
S3 as permutation matrices, we obtain the following faithful representation π :
S3 −→ GL(3, C)






1 0 0
0 1 0
1 0 0
π(1S3 ) =  0 1 0  , π((12)) =  1 0 0  , π((23)) =  0 0 1  ,
0 0 1
0 0 1
0 1 0






0 0 1
0 1 0
0 0 1
π((13)) =  0 1 0  , π((123)) =  0 0 1  , π((132)) =  1 0 0  .
1 0 0
1 0 0
0 1 0
Then it is easy to see that
χπ (1S3 ) = 3, χπ ((12)) = χπ ((13)) = χπ ((23)) = 1, χπ ((123)) = χπ ((132)) = 0.
Notice that χπ (g) is equal to the number of fixed points of g on {1, 2, 3}.
Now
< χπ , χπ > =
=
1
{1[χπ (1S3 )]2 + 3[χπ ((12))]2 + 2[χπ ((123))χπ ((132))]}
6
1
{1[9] + 3[1] + 2[0]} = 12/6 = 2.
6
this shows that χπ (and hence π) is not irreducible. Hence χπ = χi + χj , where
χi and χj are two distinct irreducible characters of S3 . (Note: If χ is a character
of a group G such that < χ, χ >= 2, then there are χi , χj ∈ Irr(G) such that
Pk
χ = χi + χj , i 6= j. Because if χ = i=1 λi χ, then < χ, χ >= 2 implies
2 =< χ, χ >=<
k
X
λi χ,
i=1
k
X
λi χ >=
i=1
k
X
λ2i .
i=1
Hence there are i 6= j for which λi = λj = 1. So that χ = χi + χj .)
Since
deg(χπ ) = 3 = χπ (1S3 ) = χi (1S3 ) + χj (1S3 ),
W.L.O.G we may assume that deg(χi ) = 1 and deg(χj ) = 2. Let us now consider
the actions of χ1 (the trivial character) and χπ on the conjugacy clases of S3 :
S3
C1
Class Rep 1S3
hi
1
χ1
1
χπ
3
C2
(12)
3
1
1
C3
(123)
2
1
0
Now
< χπ , χ1 > =
=
1
{1[χπ (1S3 )χ1 (1S3 ] + 3[χπ ((12))χ1 ((12))] + 2[χπ ((123))χ1 ((132))]}
6
1
{3 + 3 + 0} = 1.
6
5
CHARACTERS OF FINITE GROUPS
25
thus χ1 appears only once in χπ and hence χπ = χ1 + χ2 where χ2 is a nontrivial
irreducible character of S3 . Now we have
χ2 (g) = χπ (g) − χ1 (g) = χπ (g) − 1, f or all g ∈ S3 .
So that we have
χ2 (1S3 ) = 3 − 1 = 2, χ2 ((12)) = χ2 ((13)) = χ2 ((23)) = 1 − 1 = 0
and
χ2 ((123)) = χ2 ((132)) = 0 − 1 = −1.
Since |Irr(S3 )| ≤ 3, We may have one more irreducible character [in fact later
we will show that for any finite group G, |Irr(S3 )| = r, the number of conjugacy
classes if G.]. Define ρ : S3 −→ C by ρ(g) = 1 if g is even and ρ(g) = −1 if g is
odd. Then ρ is a representation of degree 1 with χρ = ρ. Notice that χρ 6= χ1 and
χρ (1S3 ) = χρ ((132)) = χρ ((123)) = 1
and
χρ ((12)) = χρ ((13)) = χρ ((23)) = −1.
Since deg(χρ ) = 1, χρ is irreducible (note that
< χρ , χρ >=
1
1
[1(1) + 3(−1)(−1) + 2(1)(1)] = [1 + 3 + 2] = 1.)
6
6
This character is the third irreducible character of S3 , namely χ3 . We are now
able to produce the following table for S3 , which is called the Character Table
of S3 over C.
Class Rep 1S3
hi
1
χ1
1
χ2
2
χ3
1
(12) (123)
3
2
1
1
0
−1
−1
1
Notice that
< χ1 , χ2 >=< χ1 , χ3 >=< χ2 , χ3 >= 0;
X
χ1 (g) = 1(1) + 3(1) + 2(1) = 6 = |S3 |;
g∈G
X
χ2 (g) = 1(2) + 3(01) + 2(−1) = 0;
g∈G
X
χ3 (g) = 1(1) + 3(−1) + 2(1) = 0.
g∈G
Let φ : S3 −→ GL(2, C) be given by
1 0
0
φ(1S3 ) =
, φ((12)) =
0 1
1
1
0
, φ((13)) =
−1
0
−1
1
,
5
CHARACTERS OF FINITE GROUPS
φ((23)) =
1
−1
0
−1
, φ((123)) =
−1
1
26
−1
0
, φ((132)) =
0
−1
1
−1
.
Then φ is a faithful representation of S3 with χφ = χ2 . Hence φ is an irreducible
representation of S3 .
Theorem 5.8 (Regular Representation) Let χπ be the character afforded by
the right regular representation of G. Let k = |Irr(G)|. Then we have
Pk
(i) χπ = i=1 χi (1G )χi ,
Pk
(ii) χπ (1G ) = i=1 [χi (1G )]2 = |G|,
Pk
(iii) χπ (g) = i=1 χi (1G )χi (g) = 0, for all g ∈ G − {1G }.
Pk
Proof: Assume that χπ = i=1 ni χi , where ni are non-negative integers. We
claim that ni = deg(χi ). We know that π(g) is a permutation on G, for all g ∈ G.
Since xg = x if and only if g = 1G , π(g) moves all the letters if g 6= 1G . Hence
χπ (g) =| G |, if g = 1G , and 0 otherwise.
Since
k
k
X
X
< χπ , χj >=<
ni χi , χj >=
ni < χi , χj >,
i=1
i=1
by the orthogonality of irreducible characters we have
< χπ , χj >= nj < χj , χj >= nj .
Thus
nj
=
< χπ , χj >=
1 X
χπ (g)χj (g −1 )
|G|
g∈G
=
1
(| G | χj (1G )) = χj (1G ) = deg(χj ), f or all j.
|G|
(i) By above
χπ =
k
X
ni χi =
i=1
k
X
χi (1G )χi .
i=1
(ii) Since χπ (1G ) =| G |, by part (i) we have
χπ (1G ) =| G |=
k
X
χi (1G )χi (1G) =
i=1
k
X
[χi (1G )]2 .
i=1
(iii) Since χπ (g) = 0 for all g ∈ G − {1G }, by part (i) we have
0 = χπ (g) =
k
X
i=1
χi (1G )χi (g).
5
CHARACTERS OF FINITE GROUPS
27
Exercise 5.4 Let A be a square matrix over a field F. Assume that for some
n ∈ N we have An = I, the identity matrix. If F contains all the nth roots of 1,
show that A is similar to a diagonal matrix.
Lemma 5.9 If ρ is a representation of G and g is an element of G, show that
there is a representation φ of G such that φ is equivalent to ρ and φ(g) is a
diagonal matrix.
Proof: Let | G |= n. Then g n = 1G , so that [ρ(g)]n = Im , where m = deg(ρ).
Since C contains all the solutions for the equation xn = 1, ρ(g) is similar to
a diagonal matrix Dg . So there is a non-singular matrix P such that Dg =
P ρ(g)P −1 . Now define φ : G → GL(m, C) by φ(h) = P ρ(h)P −1 , for all h in G.
Then φ is a representation of G equivalent to ρ with φ(g) diagonal. Definition 5.2 (Algebraic Integers) A complex number α is said to be an Algebraic Integer if it is a root of an equation of the form
xn + a1 xn−1 + a2 xn−2 + · · · + an−1 x + an = 0, ai ∈ Z.
Remark 5.1 (Algebraic Numbers) A complex number α is said to be an Algebraic Number if there is p(x) ∈ Q[x] such that P (α) = 0. It can be shown
that the set of all algebraic numbers is a subfield of C. If α is not √
an algebraic
number, then we say that it is Transcendental. For example i and 2 are algebraic numbers (in fact they are algebraic integers). Hermite, C (1822–1905) and
later Hilbert, D proved that e is transcendental. Lindemann, CLF (1852–1939)
in 1882 proved the transcendence of π. Hilbert’s 7th problem is concerned with
the transcendence of complex numbers of the form ab :
Hilbert’s Seventh Problem If a, b ∈ C such that a is an algebraic number
and a ∈
/ {0, 1}, and b is an irrational algebraic number, then ab is transcendental.
A O √Gelfond in 1934 proved that Hilbert’s seventh problem is true. For example 2 2 , 2i and ii are transcendental. But what about the case when a and b
are both transcendental? It is not known whether π π , π e or ee is transcendental.
However note that since
1
1
eπ = −π = 2i = i−2i ,
e
i
eπ is transcendental.
Now we establish some basic results on algebraic integers. In th following
we show that the set of all algebraic integers form a ring. This ring plays a
fundamental role in Number Theory.
Lemma 5.10 Let α1 , α2 , · · · , αk be complex numbers, not all zero, and suppose
that α ∈ C satisfy equations of the form
ααi =
k
X
aij αj , i = 1, 2, · · · , k,
j=1
where aij ∈ Z. Then α is an algebraic integer.
(1)
5
CHARACTERS OF FINITE GROUPS
28
Proof: Equations in (1) represents a linear homogeneous system for α1 , α2 , · · · , αk .
Since, by the hypothesis, system (1) has non-zero solution, the determinant of the
coefficient matrix must be equal to zero, that is


α − a11
−a12
−a13 · · ·
−a1k
 −a21
α − a22 −a23 · · ·
−a2k 



 = 0.
···
···
···
···
det  · · ·

 ···

···
···
···
···
−ak1
−ak2
−ak3 · · · α − akk
We can see that the above determinant is a monic polynomial of degree k in α
with integer coefficients. Hence α is an algebraic integer. Lemma 5.11 If α and β are algebraic integers, so are α + β and αβ.
Proof: Suppose that α and β satisfy the following polynomial equations
αr = a1 αr−1 + a2 αr−2 + · · · + ar−1 α + ar , ai ∈ Z,
β s = b1 β s−1 + b2 β s−2 + · · · + bs−1 β + bs , bi ∈ Z.
Then for any non-negative integer l, αl can be written as a linear combination
(with integer coefficients) of 1, α, α2 , · · · , αr−1 . Similarly for any non-negative
integer m, β m can be written as a linear combination (with integer coefficients)
of 1, β, β 2 , · · · , β s−1 .
Let α1 , α2 , · · · , αk be the products αi β j , where i.j ∈ Z, 0 ≤ i ≤ r − 1 and
0 ≤ j ≤ s − 1, arranged in some fixed order. Then any product of the form αl β m
can be represented in terms of αi β j , that is in terms of α1 , α2 , · · · , αk with integer
coefficients. Hence there are equations
(α + β)αi =
k
X
cij αj , 0 ≤ i ≤ k, cij ∈ Z,
j=1
(αβ)αi =
k
X
dij αj , 0 ≤ i ≤ k, dij ∈ Z.
j=1
Now Lemma 5.2.10 implies that α + β and αβ are algebraic integers. Exercise 5.5 Let α ∈ Q. If α is an algebraic integer, show that α ∈ Z.
Theorem 5.12 If χ is a character of a group G, then for any g ∈ G, χ(g) is an
algebraic integer.
Proof: Since G is finite, g n = 1G for some n ∈ N. Let ρ be a representation of
degree m of G that affords χ. Then [ρ(g)]n = Im , and by Lemma 5.2.9 ρ(g) is
similar to a diagonal matrix. W.L.O.G we may assume that ρ(g) itself is diagonal
(because similar matrices have the same trace). So let


1
0
0 ···
0
 0
2
0 ···
0 



ρ(g) = diag(1 , 2 , · · · , m ) =  · · · · · · · · · · · · · · · 
,
 ··· ··· ··· ··· ··· 
0
0 · · · · · · m
5
CHARACTERS OF FINITE GROUPS
29
with i ∈ C. Now since [ρ(g)]n = Im , we have ni = 1, which imply that i ’s are
nth
Pm roots of unity and hence are all algebraic integers. Since χ(g) = trace(ρ(g)) =
i=1 i , by Lemma 5.2.11 we have that χ(g) is an algebraic integer. In fact χ(g)
is a sum of nth roots of unity, where n = o(g). If bar (-) denotes the complex conjugation a + bi = a − bi in C, then we have
the following result on the conjugation of character values:
Corollary 5.13 If χ is a character of a group G, then for any g ∈ G we have
χ(g −1 ) = χ(g).
Pm
Proof: By the Theorem 5.2.12, we have χ(g) = j=1 j , where i ’s are nth roots
of unity with n = o(g) and ρ(g) similar to diag(1 , 2 , · · · , m ). Since ρ(g −1 ) =
[ρ(g)]−1 , ρ(g −1 ) is similar to
−1
−1
[diag(1 , 2 , · · · , m )]−1 = diag(−1
1 , 2 , · · · , m )
Pm
2kj π
and hence χ(g −1 ) = j=1 −1
j . We know that j = exp( n i) for some kj ∈ Z
for all
such that 0 ≤ kj ≤ n − 1. Since j j = |j |2 = 1, we deduce that j = −1
j
0 ≤ j ≤ m. Hence
χ(g −1 ) =
m
X
−1
j =
j=1
m
X
m
X
j ) = χ(g).
j = (
j=1
j=1
Exercise 5.6 Let ρ be a representation of a group G. Assume that χ is the
character afforded by ρ. Show that
(i) |χ(g)| ≤ χ(1G ), for all g ∈ G.
(ii) If |χ(g)| = χ(1G ), then ρ(g) is a scalar matrix.
(iii)* χ(g) = χ(1G ) if and only if g ∈ ker(ρ).
Definition 5.3 (F-Algebra) If F is afield and A is a vector space over F , then
we say that A is an F -Algebra if
(i) A is a ring with identity,
(ii) for all λ ∈ F and x, y ∈ A, we have λ(xy) = λ(x)y = x(λy).
Example 5.2
(i) Mn×n (F ) is the algebra of all n × n matrices over a field F .
(ii) Let V be vector space over F . Consider End(V ) = L(V, V ) = HomF (V, V ).
Then End(V ) is a ring with identity under the addition and composition
of linear transformations on V , that is (f + g)(α) = f (α) + g(α) and (f ◦
g)(α) = f (g(α)). for all α ∈ V and all f, g ∈ End(V ). Define the scalar
multiplication on End(V ) by (λf )(α) = λf (α), for all f ∈ End(V ) and
for all α ∈ V. Then End(V ) is a vector space over F. Now
[λ(f ◦ g)](α)
=
λ[(f ◦ g)(α)] = λ[f (g(α))]
=
(λf )(g(α)) = [(λf ) ◦ g](α).
5
CHARACTERS OF FINITE GROUPS
30
Hence λ(f ◦ g) = (λf ) ◦ g. Similarly we have
[λ(f ◦ g)](α)
=
λ[(f ◦ g)(α)]
=
(f ◦ g)(λα), since f ◦ g ∈ End(V )
=
f (g(λα)) = f (λg(α))
=
[f ◦ (λg)](α),
so that λ(f ◦ g) = f ◦ (λg).
Thus we have shown that End(V )is an F -Algebra.
Now we introduce another example of an algebra, the one which plays an
important part in the theory of representations, namely the Group Algebra C[G].
Definition 5.4 (Group Algebra) Let G be a finitePgroup and F be any field.
Then by F [G] we mean the set of formal forms { g∈G λg .g : λg ∈ F }. We
define the operations on F [G] by
P
P
P
(i)
g∈G λg g +
g∈G µg g :=
g∈G (λg + µg )g,
P
P
(ii) λ( g∈G λg g) := g∈G (λλg )g, λ ∈ F,
P
P
P
P
(iii) ( g∈G λg g).( g∈G µg g) := g∈G [ h∈G λh µh−1 g ]g.
Notice that the definition of multiplication given in (iii) is the result of assuming linearity and the multiplication in G. Under the above operations, F [G] is
an F-algebra. The element of F [G] for which λg = 1F and λh = 0F if h 6= g is
identified by g, that is 1F .g = g. Under this identification we embed G into F [G]
and in fact G becomes a basis for F [G].
Remark 5.2 (i) If |G| > 1, then F [G] has always zero divisors: Let g ∈ G
such that o(g) = m 6= 1. Then 1G − g and 1G + g + g 2 + · · · + g m−1 are two
non-zero elements of F [G], and we have
(1G − g).(1G + g + g 2 + · · · + g m−1 ) = 1G − g m = 1G − 1G = 0.
√
(ii) √
Consider the group G = V4 = {e, a, b, c}, F = C. let e + ia + 2 b and
2 b − ic be two elements of C[G]. Then
√
√
√
√
√
(e + ia + 2 b).( 2 b − ic) =
2 b − ic + 2 iab − i2 ac + 2b2 − i 2 bc
√
√
√
=
2 b − ic + 2 ic + b + 2e − i 2 a
√
√
√
= 2e − i 2 a + ( 2 + 1)b + i( 2 − 1)c.
Alternatively we can use part (iii) of the Definition 5.2.4, and we get
√
√
√
√
(e + ia + 2 b).( 2 b − ic) = (0 + 0 + 2 × 2 + 0)e
√
+ (1 × 0 + i × 0 + 2 × −i + 0)a
√
√
+ ( 2 + i × −i + 0 + 0)b + (−1 + 2 + 0 + 0)c.
5
CHARACTERS OF FINITE GROUPS
31
(iii) Obviously G is a subgroup of UF [G] .
Exercise 5.7 (i) Let G = D8 =< r, s : r4 = s2 = e, rs = sr−1 > . Assume
that α = r2 + r − 2s and β = −3r2 + rs are two elements of the integral
group ring Z[G]. Compute βα, αβ − βα and βαβ.
(ii) Consider the following elements of Z[S3 ]:
α = 3(1 2) − 5(2 3) + 14(1 2 3), β = 6e + 2(2 3) − 7(1 2 3),
where e = 1S3 . Compute the following elements of Z[S3 ] : 2α − 3β, βα, αβ.
Definition 5.5 (Class Sums) Let C1 , C2 , · · · , Cr be the conjugacy classes
P of elements of G. For 1 ≤ i ≤ r we define the Class Sums Ki by Ki = g∈Ci g.
Then clearly Ki ∈ F [G] and we have the following result:
Theorem 5.14 The set {K1 , K2 , · · · , Kr } is a basis for the centre of the group
ring C[G].
Proof: If g ∈ G, then g −1 Ci g = Ci . Hence we have
X
X
X
g −1 Ki g = g −1 (
h)g =
g −1 hg =
h0 = Ki .
h∈Ci
h∈Ci
h0 ∈Ci
Thus Ki g = gKi for all g ∈ G. Hence Ki ∈ Z(C[G]), where Z(C[G]) denotes the
centre of C[G].
Since distinct conjugacy classes arePdisjoint, {K1 , K2 , · · · , Kr } is a linearly
independent set (why?). Now let u = g∈G λg g be an element of Z(C[G]). Let
x ∈ G. Then
X
X
xu =
λg xg =
λg (xgx−1 )x,
g∈G
ux =
X
λg (gx) =
g∈G
g∈G
X
λxgx−1 (xgx−1 )x,
g∈G
P
and since ux = xu, we get g∈G λg (xgx−1 )x = g∈G λxgx−1 (xgx−1 )x. Hence
λg = λxgx−1 for allPx ∈ G. Thus
of all conjgates of g are the same
P the coefficients
Pr
r
in u. Hence u =
(λ
g)
=
λ
Ki . Thus {K1 , K2 , · · · , Kr } is a
i
i
i=1
g∈Ci
i=1
basis for Z(C[G]). P
Remark 5.3 Let ρ be a representation of G. Then ρ is a homomorphism from G
into GL(n, C) for some n ∈ N. We can extend ρ by linearity to an C-algebra homomorphism ρ : C[G] −→ Mn×n (C). Conversely if ρ : C[G] −→ Mn×n (C) is a representation of C[G] (that is ρ is an C-algebra homomorphism), then ρ(1G ) = In .
It follows that for all g ∈ G, ρ(g) is non-singular and [ρ(g)]−1 is equal to ρ(g −1 ).
Hence the restriction of ρ to G (note that G ⊆ C[G]) will be a representation of
G over C.
Theorem 5.15 If ρ is an irreducible representation of degree m of C[G] with the
character χ, then
5
CHARACTERS OF FINITE GROUPS
32
(i) ρ(Ki ) = di Im , di ∈ C,
(ii) Ki Kj =
(iii) di dj =
Pr
k=1
Pr
k=1
λijk Kk , λijk ∈ N ∪ {0},
λijk dk ,
(iv) di = hi χ(gi )/χ(1G ), hi = |Ci |, gi ∈ Ci .
Proof: (i) Since Ki ∈ Z(C[G]) by Theorem 5.2.14, ρ(Ki ) commutes with all
elements of ρ(G). Now since ρ is irreducible, it follows from Corollary 5.1.9 that
ρ(Ki ) = di Im for some di ∈ C.
(ii) Since Ki , Kj ∈ Z(C[G]),
Prwe have Ki Kj ∈ Z(C[G]). So by Theorem 5.2.14,
∃λijk ∈ C such that Ki Kj = k=1 λijk Kk . If we write this equation in terms of
elements of G, then since the coefficients on the left hand-side are non-negative
integers, λijk must be non-negative integers.
(iii) Using parts (i) and (ii), we get
di dj Im
= ρ(Ki )ρ(Kj ) = ρ(Ki Kj ) = ρ(
r
X
λijk Kk )
k=1
=
r
X
λijk ρ(Kk ) = (
k=1
r
X
λijk dk )Im .
k=1
Pr
Hence di dj = k=1 λijk dk .
(iv) By part (i) we have
X
X
hi χ(gi ) =
χ(g) =
trace(ρ(g))
g∈Ci
g∈Ci
= trace(
X
g∈Ci
ρ(g)) = trace(ρ(
X
g)) = trace(ρ(Ki ))
g∈Ci
= trace(di Im ) = mdi = di χ(1G ).
Hence di = hi χ(gi )/χ(1G ). Corollary 5.16 The di ’s in Theorem 5.2.15 are algebraic integers.
Pr
Proof: By part (iii) of Theorem 5.2.15, we have di dj =
k=1 λijk dk , where
λijk are non-negative integers. For a fixed j, let B be the r × r matrix (λijk )
and D be the column matrix (dk )r×1 . Then we have (dj Ir )D = BD and hence
(B − dj Ir )D = 0r×r .
Since by Theorem 5.2.15 (part (iv)) we have
d1 = h1 χ(1G )/χ(1G ) = h1 = 1 6= 0,
D is a non-zero matrix. Hence B−dj Ir is a singular matrix, so that det(B−dj Ir ) =
0. Since λijk are integers, the equation det(B − dj Ir ) = 0 produces a polynomial
5
CHARACTERS OF FINITE GROUPS
33
equation for dj with integer coefficients and leading coefficient of ±1. Thus dj is
an algebraic integer. Note: If Ci is a conjugacy class of G, then Ci0 = {g ∈ G : g −1 ∈ Ci } is also a
conjugacy class of G. Obviously Ci = Ci0 if and only if g ∼ g −1 for all g ∈ Ci .
Theorem 5.17 (Orthogonality relations) Let Irr(G) = {χ1 , χ2 , · · · , χk }. Then
P
1
(i) |G|
g∈G χi (g)χj (g) = δij , row-orthogonality.
(ii)
Pk
s=1
χs (gi )χs (gj ) = δij 0 |CG (gj )|, column-orthogonality relation.
Proof: (i)
δij
= < χi , χj >=
1 X
χi (g)χj (g −1 )
|G|
g∈G
1 X
χi (g)χj (g)
|G|
=
g∈G
by Corollary 5.2.13.
Pr
(ii) We know that Ki Kj = m=1 λijm Km . Then 1G occurs in the expansion
of Ki Kj if and only if i = j 0 (that is gi is conjugate to gj−1 ). Thus λij1 = 0 if
i 6= j 0 and λij1 = hi if i = j 0 . For each 1 ≤ s ≤ k, using Theorem 5.2.15 we get
di dj
=
=
[hi χs (gi )/χs (1G )] × [hj χs (gj )/χs (1G )]
r
X
λijm [hm χs (gm )/χs (1G )].
m=1
Thus
hi hj χs (gi )χs (gj ) =
r
X
λijm hm χs (1G )χs (gm ).
m=1
Therefore
hi hj
k
X
χs (gi )χs (gj ) =
s=1
λij1 h1
k
X
r
X
[λijm hm
m=1
χs (1G )χs (1G ) +
r
X
[λijm hm
m=2
s=1
k
X
χs (1G )χs (gm )] =
s=1
k
X
χs (1G )χs (gm )] = λij1 |G| + 0,
s=1
by Theorem 5.2.8. This show that
k
X
χs (gi )χs (gj )
=
λij1 |G|/hi hj
=
0 × |G|/hi hj = 0, if i 6= j 0
=
hi × |G|/hi hj = |G|/hj = |CG (gj )|, if i = j 0 .
s=1
Hence
Pk
s=1
χs (gi )χs (gj ) = δij 0 |CG (gj )|.
5
CHARACTERS OF FINITE GROUPS
Pk
Exercise 5.8 Show that
s=1
34
χs (gi )χs (gj ) = δij |CG (gj )|.
Theorem 5.18 (The number of irreducible characters) The number of irreducible characters of a group G equals the number of comjugacy classes of G.
Proof: Let Irr(G) = {χ1 , χ2 , · · · , χk } and let r be the number of conjugacy
classes of G. Then by the Theorem 5.2.6 we have k ≤ r. Now let
S = {(χ1 (gi ), χ2 (gi ), · · · , χk (gi )) : 1 ≤ i ≤ r}.
We claim that S is a linearly independent subset of Ck . Assume that ∃λi ∈ C
such that
r
X
λi (χ1 (gi ), χ2 (gi ), · · · , χk (gi )) = 0.
i=1
Then we must have
Pr
i=1
[
λi χs (gi ) = 0, 1 ≤ s ≤ k. So for each j we have
r
X
λi χs (gi )]χs (gj ) = 0, 1 ≤ s ≤ k.
i=1
Hence
k X
r
X
[
λi χs (gi )]χs (gj ) = 0 f or all j.
s=1 i=1
So that
r
X
i=1
k
X
λi [
χs (gi )χs (gj )] = 0 f or all j.
s=1
Now applying Theorem 5.2.17 (ii), we get
r
X
λi δij 0 |CG (gj )| = 0.
i=1
That is λj 0 |CG (gj )| = 0, so that λj 0 = 0 for all 1 ≤ j ≤ r. This shows that λj = 0
for all 1 ≤ j ≤ r. Thus S is a linearly independent subset of Ck , and hence we
have
r = |S| ≤ dim(Ck ) = k.
Therefore r = k as required. Note: Let ∆ be the r × r matrix (χi (gj )) = (aij ). Then ∆ is called the character
table of G. The rows are indexed by the irreducible characters of G and the
columns by the conjugacy classes of G. We take the first row and first column
to be indexed by the trivial character and 1G respectively, that is χ1 is the trivial
character and g1 = 1G . Theorem 5.2.18 shows that columns of ∆ are linearly
independent, and hence ∆ is non-singular. In particular the rows of ∆ are also
linearly independent.
Exercise 5.9 Compute ∆−1 . [Hint: First let B = (bjl )r×r , where bjl =
Then show that B = ∆−1 ].
1
|CG (gj )| χl (gj ).
5
CHARACTERS OF FINITE GROUPS
35
Note: Property (ii) in Theorem 5.2.17 implies that
r
X
[χs (gi )]2
=
0 if gi not conjugate to gi−1
=
|CG (gi )| otherwise.
s=1
Pr
2
In particular we have
s=1 [χs (1G )] = |G|, which is the result we proved in
Theorem 5.2.8. This shows that the sum of squares of the degrees of the irreducible
characters of G is |G|.
Exercise 5.10 Let ρ be a representation of G of degree m. Define ρ∗ from G into
GL(m, C) by ρ ∗ (g) = [ρ(g −1 )]t , transpose of ρ(g −1 ). Then show that
(i) ρ∗ is a representation of degree m of G,
(ii) χρ∗ (g) = χρ (g), for all g ∈ G,
(iii) If ρ is irreducible, so is ρ∗,
(iv) If ρ ∼ φ, then ρ∗ ∼ φ ∗ .
Theorem 5.19 The degree of an irreducible representation of a finite group G
divides |G|.
k)
are algebraic integers for all k and i. By
Proof: By Theorem 5.2.15 (iv), hχkiχ(1i (g
G)
Theorem 5.2.12, each χj (gk ) is an algebraic integer. Hence
α=
r
r X
X
j=1 k=1
hk
χi (gk )
χj (gk )
χi (1G )
is an algebraic integer by Lemma 5.2.11. Now
α
=
r X
r
X
χi (g)χj (g) X X
1
=
[χi (g)χj (g)]
χi (1G )
χi (1G )
j=1
j=1
g∈G
=
r X
X
j=1 g∈G
g∈G
1
[χi (g)χj (g −1 )]
χi (1G )
r
=
r
1 XX
|G| X
[
[χi (g)χj ∗ (g)] =
δij∗ = |G|/χi (1G ),
χi (1G ) j=1
χi (1G ) j=1
g∈G
by Theorem 5.2.17, part (i). Hence α ∈ Q. Since α is an algebraic integer and
α ∈ Q, we must have α ∈ Z. Thus χi (1G ) divides |G|. Note: The integers λijk defined in the Theorem 5.2.15 (ii) are called Class
Algebra Constants. In the following corollary we will produce a formula for
these integers. This formula plays an important role in the application of character
theory of finite groups.
5
CHARACTERS OF FINITE GROUPS
36
Corollary 5.20
r
λijk =
X χs (gi )χs (gj )χs (gk )
|G|
.
|CG (gi )||CG (gj )| s=1
χs (1G )
Proof: Let ρs denote the representation that affords χs . Since Ki Kj =
we have
r
X
ρs (Ki )ρs (Kj ) = ρs (Ki Kj ) =
λijm ρs (Km ). (∗)
Pr
m=1
λijm Km ,
m=1
Now since ρs (Ki ) = di In and ρs (Kj ) = dj In , where n is the degree of ρs (see
Theorem 5.2.15), we have
ρs (Ki ) = hi
χs (gj )
χs (gi )
In and ρs (Kj ) = hj
In ,
χs (1G )
χs (1G )
by Theorem 5.2.15. Now by using the relation (*) above, we get
hi
Hence
r
X
r
X
χs (gj )
χs (gm )
χs (gi )
× hj
=
λijm hm
.
χs (1G )
χs (1G ) m=1
χs (1G )
λijm hm χs (gm ) = hi hj χs (gi )χs (gj )/χs (1G ),
(1)
m=1
multiplying by both sides of (1) by χs (gk ) and summing from s = 1 to s = r we
obtain
r
r
X
X
χs (gi )χs (gj )χs (gk )
χs (gm )χs (gk )] = hi hj
λijm hm [
.
χs (1G )
s=1
m=1
s=1
r
X
Since
r
X
χs (gm )χs (gk ) = δkm |CG (gk )|
s=1
by Exercise 5.2.7, we have
r
X
λijm hm δkm |CG (gk )| = hi hj
m=1
r
X
χs (gi )χs (gj )χs (gk )
s=1
So that
λijk hk |CG (gk )| = hi hj
r
X
χs (gi )χs (gj )χs (gk )
χs (1G )
s=1
Thus
r
λijk |G| =
χs (1G )
.
X χs (gi )χs (gj )χs (gk )
|G||G|
.
|CG (gi )||CG (gj )| s=1
χs (1G )
This gives the desired formula for λijm .
.
(2)
5
CHARACTERS OF FINITE GROUPS
37
Example 5.3 (i) If g 2 = 1G , then χi (g) ∈ Z for all χi ∈ Irr(G) : Because
g 2 = 1G implies that g = g −1 . If g = 1G , then χi (g) = χi (1G ) = deg(χi ) ∈ Z. If g
is not the identity, then o(g) = 2 and hence χi (g) is a sum of 2nd roots of unity.
Since the roots are 1 and -1, clearly χi (g) ∈ Z.
(ii) If g is conjugate to g −1 , then χi (g) ∈ R for all χi ∈ Irr(G) : Because g
conjugate to g −1 implies that χi (g) = χi (g −1 ) = χi (g). Thus χi (g) ∈ R.
(iii) Assume that g ∈ G is an element of order three and g ∼ g −1 . Then χi (g) ∈
Z for all χi ∈ Irr(G) : Because g ∼ g −1 implies that χi (g) ∈ R for √all χi ∈ Irr(G),
√
by part (ii). Let χi (g) = ε1 + ε2 + · · · + εm , where i ∈ {1, − 12 − 23 i, − 21 + 23 i}.
Assume that for some j, 1 ≤ j ≤ m, we have
√
√
3
3
1
1
i or j = − +
i.
j = − −
2
2
2
2
Then since χi (g) ∈ R, j must also appear in χi (g). Now since j + j = −1, we
deduce that χi (g) ∈ Z.
Example 5.4 (Character Table of S4 ) In S4 there are 5 conjugacy classes
and hence Irr(S4 ) = 5. Consider the map ρ2 : S4 −→ C given by ρ2 (α) = 1
if α is even and ρ2 (α) = −1 if α is odd. Then ρ2 is a representation of degree 1
and hence ρ2 = χρ2 . let denote this character by χ2 . Then we have
χ2 (1S4 ) = χ2 ((1 2)(3 4)) = χ2 ((1 2 3)) = 1
and
χ2 ((1 2)) = χ2 ((1 2 3 4)) = −1.
So we have the following table:
Classes of S4
hi
χ2
1S4
1
1
(1 2)(3 4)
3
1
(1 2)
6
−1
(1 2 3 4)
6
−1
(1 2 3)
8
1
Since
< χ2 , χ2 > =
=
1
[1 + 3(1)(1) + 6(−1)(−1) + 6(−1)(−1) + 8(1)(1)]
24
1
[1 + 3 + 6 + 6 + 8] = 24/24 = 1,
24
χ2 is irreducible.
Now let π : S4 −→ GL(4, C) be the natural permutation representation of S4 .
Then χπ (g) is equal to the number of fixed points of g on the set {1, 2, 3, 4}. Then
we have
Classes of S4
hi
χπ
1S4
1
4
(1 2)(3 4)
3
0
(1 2)
6
2
(1 2 3 4)
6
0
(1 2 3)
8
1
It is not difficult to see that < χπ , χπ >= 2 and < χπ , χ1 >= 1, where χ1 is the
trivial charascter. hence χπ = χ1 + χ3 , where χ3 is an irreducible character of
degree 4 − 1 = 3. Then we have χ3 (g) = χπ (g) − χ1 (g), for all g in S4 .
5
CHARACTERS OF FINITE GROUPS
38
Now itP
remains to find two more irreducible characters of S4 , namely χ4 and
5
χ5 . Since i=1 [χi (1S4 )]2 = |G| = 24, we have
[χ4 (1S4 )]2 + [χ5 (1S4 )]2 = 24 − (1 + 1 + 9) = 24 − 11 = 13 = 4 + 9.
This implies that we can assume deg(χ4 ) = 2 and deg(χ4 ) = 3. So far we have
the following information for the character table of S4 :
1S4
1
1
1
3
2
3
Classes of S4
hi
χ1
χ2
χ3
χ4
χ5
(1 2)(3 4)
3
1
1
−1
a
e
(1 2) (1 2 3 4)
6
6
1
1
−1
−1
1
−1
b
c
f
g
(1 2 3)
8
1
1
0
d
h
We are able to complete the character table by means of the orthogonality relations. First notice that, since g ∼ g −1 for all g ∈ S4 , we have {a, b, c, d, e, f, g, h} ⊆
R. Using Example 5.2.3, parts (i) and (ii), we have {a, e, b, f, d, h} ⊆ Z. Now using
the orthogonality of the first two columns we get
1 + 1 − 3 + 2a + 3e = 0,
so that
2a + 3e = 1.
Since
P5
i=1 [χi ((1
(1)
2
2)(3 4))] = |CS4 ((1 2)(3 4))|, by Note 5.2.7, we have
1 + 1 + 1 + a2 + e2 =
24
= 8,
3
and hence
a2 + e2 = 5.
(2)
Using relations (1) and (2) we obtain a = 2 and e = −1.
Similarly the orthogonality of the first column with columns three and five
give
1 − 1 + 3 + 2b + 3f = 0, 1 + 1 + 0 + 2d + 3h = 0.
We deduce that
2b + 3f = −3
(3)
2d + 3h = −2.
(4)
and
Since
and
5
X
24
[χi ((1 2))]2 = |CS4 ((1 2))| =
=4
6
i=1
5
X
i=1
[χi ((1 2 3))]2 = |CS4 ((1 2 3))| =
24
= 3,
8
5
CHARACTERS OF FINITE GROUPS
39
we get
b2 + f 2 = 4 − 3 = 1
(5)
d2 + h2 = 3 − 2 = 1.
(6)
and
Now relations (3) and (5) imply that b = 0 and f = −1. Similarly relations (4)
and (6) imply that d = −1 and h = 0. At this stage we produce the following
information on the character table of S4 :
Classes of S4
hi
χ1
χ2
χ3
χ4
χ5
1S4
1
1
1
3
2
3
(1 2)(3 4)
3
1
1
−1
2
−1
(1 2) (1 2 3 4)
6
6
1
1
−1
−1
1
−1
0
c
−1
g
(1 2 3)
8
1
1
0
−1
0
Using the orthogonality of columns 3 and 4 we obtain
1+1−1+0×c−g =0
and hence g = 1. Now the orthogonality of columns 4 and 5 gives
1 − 1 + (−1) × 0 + c(−1) + 1 × 0 = 0,
and hence c = 0. This completes the character table of S4 :
Classes of S4
hi
χ1
χ2
χ3
χ4
χ5
1S4
1
1
1
3
2
3
(1 2)(3 4)
3
1
1
−1
2
−1
(1 2) (1 2 3 4)
6
6
1
1
−1
−1
1
−1
0
0
−1
1
(1 2 3)
8
1
1
0
−1
0
Exercise 5.11 (Characters of cyclic groups) . Let G = hxi be a cyclic group
of order n. Let e2kπi/n be the nth roots of unity in C, k = 0, 1, 2, ..., n − 1. Define ρk : G → C∗ by ρk (xm ) = [e2kπi/n ]m . Show that ρk define the n distinct
irreducible representations of G.
Exercise 5.12 Use the above exercise to construct the character table of the
cyclic groups of order 2, 3, 4. 5 and 6.
Exercise 5.13 Calculate the character table of the V4 , the Klien 4-group.
Note: If G is an abelian group, then all irreducible representations of G are of
degree 1. In general, we would like to know how many of the irreducible representations of an arbitrary group G are of degree 1. In the following theorem we give
the answer to this question.
5
CHARACTERS OF FINITE GROUPS
40
Theorem 5.21 Let G be a finite group. The number of representations of G of
degree 1 is equal to [G : G0 ].
Proof: If ρ is a representation of degree one of G, then by Exercise 5.3(i) we
have Ker(ρ) ⊇ G0 . Now the define φ : G/G0 → C∗ by φ(gG0 ) = ρ(g), for all
g ∈ G. Then φ defines a representation of degree one for the group G/G0 (note
that since G0 ⊆ Ker(ρ), φ is well-defined.) Since G/G0 is abelian, it has [G:G0 ]
conjugacy classes. Hence the group G/G0 has [G:G0 ] irreducible characters. Since
G/G0 is abelian, all its irreducible characters are of degree one (see Exercise 5.3,
part (ii).) Now consider the natural homomorphism π : G → G/G0 . If ψ is a
representation of G/G0 of degree one, then ψ ◦ π is a representation of degree
one of G. Now it is not difficult to see that we have a one-to-one correspondence
between the set of all representations of degree one of G and the set of all the
irreducible representations of the group G/G0 .
Exercise 5.14 Compute the character table of A4 .
Exercise 5.15 Calculate the character tables of Q (the quaternion group of order
8) and D8 . Show that they have same character tables.
Exercise 5.16 Calculate the character table of A5 . Recall that A5 is a nonabelain simple group of order 60 and hence (A5 )0 = A5.
Note: In the Exercises 5.1.1 and 5.1.2 we observed that there is a one-to-one correspondence between representations of G/N and representations of G with kernel
containing N. Furthermore it is not difficult to show that, under this correspondence, irreducible representations correspond to irreducible representations. We
put this result in terms of characters in the following theorem. If χ is a character
afforded by a representation ρ of G, we define ker(χ) to be ker(ρ).
Theorem 5.22 Let N G.
(i) If χ is a character of G and N ker(χ), then χ is constant on cosets of N
in G and χ
b on G/N defined by χ
b(gN ) = χ(g) is a character of G/N.
(ii) If χ
b is a character of G/N, then the function χ defined by χ(g) = χ
b(gN ) is
a character of G.
(iii) In both (i) and (ii), χ ∈ Irr(G) iff χ
b ∈ Irr(G/N ).
5
CHARACTERS OF FINITE GROUPS
41
Proof: (i) and (ii) follow from Exercise 5.1.1 and 5.1.2.
(iii) Let S be a set of coset representatives of N in G. Then we have
1 = hχ, χi =
1 X
χ(g) · χ(g −1 )
|G|
g∈G
=
1 X
|N | · χ(g) · χ(g −1 )
|G|
=
1 X
|N | · χ
b(gN ) · χ
b(g −1 N )
|G|
=
1
|G|
=
1
|G/N |
g∈S
g∈S
X
|N | · χ
b(gN ) · χ
b(gN )−1
gN ∈G/N
X
χ
b(gN ) · χ
b(gN )−1 = hb
χ, χ
bi .
gN ∈G/N
Example 5.5 Consider the group G = S4 . Let N = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
ci = [gi N ] is a class of G/N. However, distinct
G. If Ci = [gi ] is a class of G, then C
classes in G may produce equal classes in G/N. Referring to the character table
of S4 (See Example 5.2.4), we see that
{χ| χ ∈ Irr(G), N ⊆ ker(χ)} = {χ1 , χ2 , χ4 }.
Hence Irr(G/N ) = {b
χ1 , χ
b2 , χ
b4 }. Using the character table of S4 we have
Classes of S4
χ1
χ2
χ2
1S4
1
1
2
(1 2)(3 4)
1
1
2
(1 2)
1
−1
0
(1 2 3 4)
1
−1
0
(1 2 3)
1
1
−1
Observe that columns 1 and 2 are identical, as are columns 3 and 4. Deleting
repeats, we obtain the character table of G/N
Classes of G/N
χ
b1
χ
b2
χ
b4
N
1
1
2
(1 2)N
1
−1
0
(1 2 3)N
1
1
−1
we can see that G/N is a group of order 6 and it is not abelian. Hence G/N ∼
= S3 .
(See the character table of S3 in the Example 5.2.1)
Note: If N G, then the character table of G determines whether or not G/N
is abelian. There is no way to determine from the character table of G whether
or not N is abelian.
Corollary 5.23 Let g ∈ G and N G. Then |CG (g)| ≥ |CG/N (gN )|.
6
GROUP ACTIONS AND PERMUTATION CHARACTERS
42
Proof: We know that
Irr(G/N )
=
|CG/N (gN )| =
{b
χ| χ ∈ Irr(G), N ⊆ ker(χ)}.
X
χ
b(gN ) · χ
b(gN )−1
χ
b∈Irr(G/N )
=
X
b(gN ) =
χ
b(gN ) · χ
≤
|b
χ(gN )|2
χ
b∈Irr(G/N )
χ
b∈Irr(G/N )
=
X
X
{|χ(g)|2 | χ ∈ Irr(G), N ⊆ ker(χ)}
X
|χ(g)|2 = |CG (g)|.
χ∈Irr(G)
6
Group Actions and Permutation Characters
Suppose that G is a finite group acting on a finite set Ω. For α ∈ Ω, the stabilizer
of α in G is given by
Gα = {g ∈ G|αg = α}.
Then Gα ≤ G and [G : Gα ] = |∆|, where ∆ is the orbit containing α.
The action of G on Ω gives a permutation representation π with corresponding
permutation character χπ denoted by χ(G|Ω). Then from elementary representation theory we deduce that
Lemma 6.1 (i) The action of G on Ω is isomorphic to the action of G on the
G/Gα , that is on the set of all left cosets of Gα in G. Hence χ(G|Ω) =
χ(G|Gα ).
(ii) χ(G|Ω) = (IGα )G , the trivial character of Gα induced to G.
(iii) For all g ∈ G, we have χ(G|Ω)(g) = number of points in Ω fixed by g.
Proof: For example see Isaacs [12] or Ali [1]. In fact for any subgroup H ≤ G we have
χ(G|H)(g) =
k
X
|CG (g)|
,
|CH (hi )|
i=1
where h1 , h2 , ..., hk are representatives of the conjugacy classes of H that fuse to
[g] = Cg in G.
Lemma 6.2 Let H be a subgroup of G and let Ω be the set of all conjugates of
H in G. Then we have
(i) GH = NG (H) and χ(G|Ω) = χ(G|NG (H).
6
GROUP ACTIONS AND PERMUTATION CHARACTERS
43
(ii) For any g in G, the number of conjugates of H in G containing g is given
by
χ(G|Ω)(g) =
m
X
i=1
k
X
|CG (g)|
|CG (g)|
= [NG (H) : H]−1
,
|CNG (H) (xi )|
|C
H (hi )|
i=1
where xi ’s and hi ’s are representatives of the conjugacy classes of NG (H)
and H that fuse to [g] = Cg in G, respectively.
Proof:
(i)
GH = {x ∈ G|H x = H} = {x ∈ G|x ∈ NG (H)} = NG (H).
Now the results follows from Lemma 6.1 part (i).
(ii) The proof follows from part (i) and Corollary 3.1.3 of Ganief [11] which uses
a result of Finkelstien [9]. Remark 6.1 Note that
χ(G|Ω)(g) = |{H x : (H x )g = H x }| = |{H x |H x
−1
gx
= H} =
|{H x |x−1 gx ∈ NG (H)}| = |{H x |g ∈ xNG (H)x−1 }| = |{H x |g ∈ (NG (H))x }|.
Corollary 6.3 If G is a finite simple group and M is a maximal subgroup of G,
then number λ of conjugates of M in G containing g is given by
χ(G|M )(g) =
k
X
|CG (g)|
,
|CM (xi )|
i=1
where x1 , x2 , ..., xk are representatives of the conjugacy classes of M that fuse to
the class [g] = Cg in G.
Proof: It follows from Lemma 6.2 and the fact that NG (M ) = M. It is also a
direct application of Remark 1, since
χ(G|Ω)(g) = |{M x |g ∈ (NG (M ))x }| = |{M x |g ∈ M x }|. Let B be a subset of Ω. If B g = B or B g ∩ B = ∅ for all g ∈ G, we say B
is a block for G. Clearly ∅, Ω and {α} for all α ∈ Ω are blocks, called trivial
blocks. Any other block is called non-trivial. If G is transitive on Ω such that
G has no non-trivial block on Ω, then we say G is primitive. Otherwise we say
G is imprimitive.
Remark 6.2 Classification of Finite Simple Groups (CFSG) implies that no 6transitive finite groups exist other than Sn (n ≥ 6) and An (n ≥ 8), and that the
Mathieu groups are the only faithful permutation groups other than Sn and An
providing examples for 4- and 5-transitive groups.
7
DESIGNS
44
Remark 6.3 It is well-known that every 2-transitive group is primitive. By using
CFSG, all finite 2-transitive groups are known.
The following is a well-known theorem that gives a characterisation of primitive
permutation groups. Since by Lemma 6.1 the permutation action of a group G
on a set Ω is equivalent to the action of G on the set of the left cosets G/Gα ,
determination of the primitive actions of G reduces to the classification of its
maximal subgroups.
Theorem 6.4 Let G be transitive permutation group on a set Ω. Then G is
primitive if and only if Gα is a maximal subgroup of G for every α ∈ Ω.
Proof: See Rotman [37]. 7
Designs
An incidence structure S = (P, B, I) consists of two disjoint sets P (called points)
and B (called blocks), and I ⊆ P × B. If (p, B) ∈ I, then we say that the point p
is incident with the block B. The pair (p, B) is called a flag. If (p, B) ∈
/ I, then
it is an anti-flag.
Example 7.1 Let P be any set and B ⊆ 2P , where 2P is the set of all subsets
of P (power set). Let I = {(p, B) : p ∈ P, B ∈ B, p ∈ B}. Then we have an
incidence structure S = (P, B, I).
For example let P = {1, 2, 3}, B = {{1}, {1, 2}, {2, 3}}. Then
I = {(1, {1}), (1, {1, 2}), (2, {1, 2}), (2, {2, 3}), (3, {2, 3})}.
We have three points and three blocks. Note that in this case I $ P × B.
Definition 7.1 (t-Design) An incidence structure D = (P, B, I), with point set
P, block set B and incidence I is a t-(v, k, λ) design, if |P| = v, every block B ∈ B
is incident with precisely k points, and every t distinct points are together incident
with precisely λ blocks.
We will say that a design D is symmetric if it has the same number of points
and blocks. A t − (v, k, 1) design is called a Steiner System. A 2 − (v, 3, 1)
Steiner system is called a Steiner Triple System.
A t − (v, 2, λ) design D can be regarded as a graph with P as points and B as
edges.
Example 7.2 . Consider Example 7.1, where P = {1, 2, 3}, B = {{1, 2}, {1, 3}, {2, 3}},
and (p, B) ∈ I if and only if p ∈ B. Then the design D is a 1 − (3, 2, 2) design,
which is also a 2 − (3, 2, 1) design. It is also symmetric.
Exercise 7.1 Let P = {1, 2, 3}. Consider Example 7.1 and find two more tdesigns.
7
DESIGNS
45
Remark 7.1 A Steiner system 2 − (n2 + n + 1, n + 1, 1) is called a Projective
Plane of order n. A Steiner system 2 − (n2 , n, 1) is called an Affine Plane of
order n. Projective and affine planes of order n = pk , where p is a prime, exist.
But the question is: Is there a finte plane of order n when n is not a
prime? The conjecture is that the answer is NO, but so far has not been proven.
It can be shown that a projective plane of order n exist if and only if there exits
an affine plane of order n.
Bruck-Ryser Theorem (1949)[6] states that: if n = 4k + 1 or 4k + 2 and n is
not equal to the sum of two squares of integers, then there is no projective plane
of order n.
Note that 10 is not a prime, 10 = 4 × 2 + 2, but 10 = 32 + 12 . So we cannot
use Bruck-Ryser Theorem to show the nonexistence of a finite plane of order 10.
The non-existence was proved (using computers) by Lam in 1991 (see [27]) after
two decades of search for a solution to the problem.
The next smallest number to look at is 12. We do not yet know whether there
exists a finite plane of order 12.
Figure 1: Fano Plane
Example 7.3 Fano Plane . The Fano plane is a projective plane of order 2,
which is a 2 − (7, 3, 1) design (a Stiener triple system on 7 points).
In the Figure 1 we have P = {1, 2, 3, 4, 5, 6, 7}, B = {B1 , B2 , B3 , B4 , B5 , B6 , B7 },
where B1 = {1, 2, 3},B2 = {1, 5, 6}, B3 = {1, 4, 7}, B4 = {2, 4, 6}, B5 = {2, 5, 7},
B6 = {3, 6, 7} and B7 = {3, 4, 5}.
We can see that the Fano plane is a symmetric 2-design. Also note that it is
a 1 − (7, 3, 3) design.
Remark 7.2 (Counting Principles) In combinatorics often we need to count
the number of elements of a set S in two different ways and then equate our
7
DESIGNS
46
answers. So in general assume that X and Y are two finite sets and S ⊆ X × Y.
We define
S(a, ) = {(x, y) : (x, y) ∈ S, x = a}, S( , b) = {(x, y) : (x, y) ∈ S, y = b}.
Then
S=
[
˙
S(a, ) =
[
˙
S( , b).
Hence we have
|S| =
X
|S(a, )| =
a∈X
X
|S( , b)|,
b∈Y
and if |S(a, )| = l and |S( , b)| = m are independent of a and b respectively, then
we have
l|X| = m|Y |.
We use the Counting Priciple described above (see Remark 7.2 to prove the
following theorem on t-designs.
Theorem 7.1 If D is a t − (v, k, λ) design and 1 ≤ s ≤ t, then D is also a
s − (v, k, λs ) design where
λs =
(v − s)(v − s − 1) · · · (v − t + 1)
.
(k − s)(k − s − 1) · · · (k − t + 1)
Proof: Let S be a set of s points and let m be the number of blocks that contain
S. Let
T = {(T, B) : S ⊂ T ⊂ B, |T | = t, B ∈ B}.
Now count the number of elements of T in two different way, we have
λ(
v−s
k−s
) = m(
).
t−s
t−s
We can see that m is independet of S and hence
λs = m = λ(
v−s
k−s
)/(
),
t−s
t−s
which gives the formula. Note that Fano Plane is a 2-(7,3,1) design. Here λ = λt = λ2 = 1, and hence
7−1
= 3. We deduce that Fano plane is also a 1 − (7, 3, 3) design.
λ1 = 1 × 3−1
Remark 7.3
1. λt = λ and λs =
v−s
k−s
× λs+1 .
2. If the number of blocks in a t − design D is denoted by b, then we have
b = λ0 =
v(v − 1) · · · (v − t + 1)
.
k(k − 1) · · · (k − t + 1)
If we denote λ1 (replication number) by r, then we have
r = λ1 =
(v − 1)(v − 2) · · · (v − t + 1)
.
(k − 1)(k − 2) · · · (k − t + 1)
7
DESIGNS
47
Thus we get
b=
v
r
k
and hence we deduce that
bk = vr.
3. In a 2-design 2 − (v, k, λ), we have λ2 = λ and by part (1) we get
λ1 =
v−1
× λ2 ,
k−1
and hence
λ1 (k − 1) = λ(v − 1)
so that
r(k − 1) = λ(v − 1).
Definition 7.2 (Incidence Matrix) Let D = (P, B, I) be a design in which
P = {p1 , p2 , · · · , pv } and B = {B1 , B2 , · · · , Bb }. Then the incidence matrix of D
is defined to be a b × v matrix A = (aij ) such that
aij = {
1 if (pj , Bi ) ∈ I
0 if (pj , Bi ) ∈
/I
Theorem 7.2 Let D is a 2 − (v, k, λ) design with A as its incidence matrix. If
Iv is the v × v identity matrix and Jv is the v × v matrix with all entries equal to
1, then we have
At A = (r − λ)Iv + λJv ,
and
det(At A) = (r − λ)v−1 (vλ − λ + r) = (r − v)v−1 rk.
P
Proof: Easy to see that (At A)ij = aki akj , which is equal to the inner product
of ith column of A with jth column of A. If i = j then this number is r and if
i 6= j it is λ. Hence


r λ ··· λ
 λ r ··· λ 


= (r − λ)Iv + λJv .
At A =  . . .
. . ... 
 .. ..

λ
λ
···
r
Subtract the first column from each other column, and then add each row to the
first row. We get a lower triangular matrix with diagonal entries r + (v − 1)λ, r −
λ, r − λ, · · · , r − λ. Thus
det(At A) = (r − λ)v−1 (vλ − λ + r)
and since by Remark 3 (3) vλ − λ = r(k − 1), we have
det(At A) = (r − λ)v−1 (r(k − 1) + r) = (r − λ)v−1 rk.
7
DESIGNS
48
In a t-design D, where t ≥ 2, the order of D is defined to be n = λ1 − λ2 . So
if t = 2, then n = r − λ and
det(At A) = (r − λ)v−1 rk = nv−1 rk.
Since by Theorem 7.1 any t-design with t ≥ 2 is also a 2-design, Theorem 7.2
is true for any t-design with t ≥ 2.
Corollary 7.3 If D is a none-trivial 2-design with an incedence matrix A, then
rank(A) over Q is v.
Proof: At A is an square v × v matrix and since D is non-trivial, v − k 6= 0 so
that r − λ 6= 0, (because in a 2-design we have r(k − 1) = λ(v − 1)). Thus 0 6=
det(At A) = (det(A))2 , which implies that det(A) 6= 0. Therefore rankQ (A) = v.
Example 7.4 (Incidence Matrix of Fano Plane) Consider the Example 7.3.
If we let M be the incidence matrix of the Fano plane, then we have that


1 1 1 0 0 0 0
 1 0 0 0 1 1 0 


 1 0 0 1 0 0 1 



M =
 0 1 0 1 0 1 0 .
 0 1 0 0 1 0 1 


 0 0 1 0 0 1 1 
0 0 1 1 1 0 0
Since Fano plane is a non-trivial 2-design, by Theorem 7.2 we have that rankQ (M ) =
7 and detQ (M ) = 24. Why?
Exercise 7.2 If M is the incidence matrix of the Fano plane show that
i. rankF (M ) = 7, where F is a field of characteristic p with p ∈
/ {2, 3}.
ii. rankF (M ) = 4, where char(F ) = 2;
iii. rankF (M ) = 6, where char(F ) = 3.
Corollary 7.4 If D is a t − design with t ≥ 2, then b ≥ v, that is the number of
blocks is at least same as the number of points.
Proof: Since D is a non-trivial 2-design, rankQ (A) = v. Now since A is a b × v
matrix, its rank must be less than or equal to number of its row, that is we have
v = rankQ (A) ≤ b. Definition 7.3 An isomorphim between two designs D1 and D2 is a bijection
φ between sets of points P 1 and P 2 and between sets of blocks B 1 and B 2 such that
for any p ∈ P 1 and B ∈ B 1 , pI 1 B implies that φ(p)I 2 φ(B). If D1 = D2 , then φ
is called an automorphism (or a collineation). The group of automorphisms of
a design D is denoted by Aut(D).
8
CODES
49
Example 7.5 If D is the 2-design describing the Fano plane (see Example 7.3),
then
G = Aut(D) =< (5 6)(7 4), (1 2)(6 7), (5 7)(1 3), (1 4 5 7 3 2 6) > .
The group G is 2-transitive on points with
|G| = 7 × 6 × |G12 | = 168,
where
G12 = {e, (5 6)(7 4), (5 7)(4 6), (5 4)(6 7)} ∼
= V4
∼ S4 and G ∼
Also note that G1 =
= P SL(2, 7) ∼
= P SL(3, 2).
Definition 7.4
i. The complement of D is the structure D̃ = (P, B, Ĩ),
where Ĩ = P ×B −I. If D is a t−(v, k, λ) design, then D̃ is a t−(v, v −k, λ̃)
design.
ii. The dual structure of D is Dt = (B, P, I t ), where (B, P ) ∈ I t if and only if
(P, B) ∈ I. Thus the transpose of an incidence matrix for D is an incidence
matrix for Dt . We say D is self dual if it is isomorphic to its dual.
iii. A t-(v, k, λ) design is called self-orthogonal if the block intersection numbers have the same parity as the block size.
Exercise 7.3 Show that the complement of the Fano plane is a 2−(7, 4, 2) design.
If M̃ is the incedence matrix of this complement, show that det(M̃ ) = 210 . If F
is a field of characteristic 2, show that rankF (M̃ ) = 3.
8
Codes
Definition 8.1 Let F be a finite set of q elements. A q-ary code C is a set of
code words (x1 , x2 , . . . , xn ), xi ∈ F, n ∈ N. If all code words have same length n,
then we say that C is a block code of length n. In this case C ⊆ F n .
Definition 8.2 (Hamming distance) If w = (w1 , w2 , . . . , wn ) and v = (v1 , v2 , . . . , vn )
are in F n , we define the Hamming distance d(v, w) by
d(v, w) = |{i : vi 6= wi }k.
For example in F 4 , where F = GF (3), if v = (1, 1, 2, 0) and w = (0, 1, 2, 3). then
d(v, w) = 2. The following properties of Hamming distance are easy to prove:
1. d(v, w)= 0 if and only if v = w;
2. d(v, w) = d(w, v), for all v, w ∈ F ;
3. d(u, w) ≤ d(u, v) + d(v, w), for all u, v, w ∈ F.
Definition 8.3 (Minimum Distance) If C is a code we define the minimum
distance d(C) by
d(C) = min{d(v, w) : v, w ∈ C, v 6= w}.
8
CODES
50
The following results plays an important role in detecting and correcting errors
when codes are transmitted via symmetric q-ary channels.
Theorem 8.1 Let C be a code with minimum distance d.
i. If d ≥ s + 1 ≥ 2, then C can be used to detect up to s errors.
ii. if d ≥ 2t + 1, then C can be used to correct up to t errors.
Proof: i. Suppose v is transmitted and w received with less than or equal s
errors. Then d(v, w) ≤ s ≤ d − 1 < d and hence w ∈
/ C or w = v. Thus if we had
errors, it would be detected.
ii. Suppose v is transmitted and w received with less than or equal t errors.
Then we have d(v, w) ≤ t. Now suppose that u ∈ C such that u 6= v, then we
have
d(u, w) + d(w, v) ≥ d(u, v) ≥ d ≥ 2t + 1,
and hence
d(u, w) ≥ 2t + 1 − d(v, w) ≥ 2t + 1 − t = t + 1.
Thus v is the closest codeword to w in C and it could be picked. Corollary 8.2 If d = d(C), then C can detect at most d − 1 errors and correct
at most b d−1
2 c errors.
Proof: Folows from Theorem 8.1. Theorem 8.3 (Singleton Bound) Let C be a q-ary code of length n and mininmum distance d. Then |C| ≤ q n−d+1 .
Proof: We know that C ⊆ F n and hence clearly |C| ≤ q n . Let C 0 be the set of
all code words of C that their last d − 1 co-ordinates are removed. Then clearly
|C| = |C 0 |, since all elements of C 0 are distinct due to the fact that no two code
words of C differed in less than d places. Now each cowords in C 0 has length
n − d + 1 and hence
|C| = |C 0 | ≤ q n−d+1 ,
and thus the result. 8.1
Linear Codes
From now on we regard F as a finite field Fq = GF (q) and our codes C to be
subspaces of V = F n . If dim(C) = k and d(C) = d, then the code C is denoted
by [n, k, d]q to represent this information.
Definition 8.4 (Support and Minimum Weight) Let V = F n , v = (x1 , x2 , x3 . . . , xn ) ∈
V and S = {i : vi 6= 0}. Then S is called the support of v (denoted by supp (v)),
the |S| is said to be the weight of v (denoted by wt (v)) If C is a linear code, the
minimum weight of C is min{wt(c) : c ∈ C}.
Proposition 8.4 Let C = [n, k, d]. Then we have
8
CODES
51
i. d = d(C) is the minimum weight of C,
ii. d ≤ n − k + 1.
Proof:
i. Clearly in V = F n we have d(v, w) = wt(v − w). Now since C is a subspace
of V , for any v, w ∈ C we have v − w ∈ C and hence the result.
ii. Since C is a subspace of V with dim(C) = k, we have |C| = q k and now by
Theorem 8.3 we have q k ≤ q n−d+1 . Hence k ≤ n−d+1, so that d ≤ n−k+1.
A constant word in a code is a codeword that is a scalar multiple of vector
all of whose coordinate entries are either 0 or 1.
The all-one vector will be denoted by , and is the constant vector of weight
the length of the code.
Two linear codes of the same length and over the same field are equivalent if
each can be obtained from the other by permuting the coordinate positions and
multiplying each coordinate position by a non-zero field element.
They are isomorphic if they can be obtained from one another by permuting
the coordinate positions.
An automorphism of a code is any permutation of the coordinate positions
that maps codewords to codewords. An automorphism thus preserves each weight
class of C.
A binary code with all weights divisible by 4 is said to be a doubly-even
binary code.
Definition 8.5 (Dual of a Code) For any code C, the dual code C ⊥ is the
orthogonal subspace under the standard inner product. That is
C ⊥ = {v ∈ F n :< v, c >= 0 for all c ∈ C}.
If C ⊆ C ⊥ , then we say C is self-orthogonal. If C = C ⊥ , then we say that C
is self-dual. The hull of C is given by Hull(C) = C ∩ C ⊥ .
Definition 8.6 (Generating Matrix) . If C is a q-ary code of dimension k
and of length n, then a generating matrix G for C is a k × n matrix obtained from
any basis of C.
By performing elementary row operations on G, we can reduce it into a row
echelon form G0 = [Ik |A] (standard form), where A is a k × (n − k) matrix.
Clearly G0 is a generating matrix for a code which is equivalent (and isomorphic)
to C.
Proposition 8.5 If C is a [n, k] code, then C ⊥ is a [n, n − k] code.
Proof: Let G be a generating matrix for C. Then (v)Gt ∈ F k for all v ∈ F n
and Gt can be regarded as a linear transformation from F n onto F k . Clearly
ker(Gt ) = C ⊥ and hence F n /C ⊥ ∼
= F k , that is dim(F n ) − dim(C ⊥ ) = dim(F k ).
⊥
Hence n = dim(C) + dim(C ). 8
CODES
52
Exercise 8.1 Let C =<  > be the code generated by all one vector  inside F n .
1. What is the standard form for C?
2. Show that C ⊥ is a [n, n − 1, 2] code.
3. Show that C ⊥ is generated by the vectors with exactly 2 non-zero entries 1
and −1.
4. Show that the standard form for C ⊥ is [In−1 |B], where B is the column
matrix with −1 entries.
Definition 8.7 (Parity Check Matrix) If C is a code, then any generating
matrix for C ⊥ is said to be a parity-check matrix for C.
Proposition 8.6 If G and H are generating and parity-check matrices of a code
C, then we have GH t = 0k×(n−k) and c ∈ C if and only if cH t =0 if and only if
Hct = 0.
Proof: Follows from the fact that C and C ⊥ are orthogonal and H is a generating matrix for C ⊥ . Note that if G = [Ik |A] is a generating matrix for a code C of dimension k in F n ,
then H = [−At |In−k ] is a parity-check matrix for C. A generating matrix in its
standard form simplifies the encoding. For example if we encode u ∈ F k by G =
[Ik |A], we compute uG and we will get v = (u1 , u2 , . . . , uk , xk+1 , xk+2 . . . , xn ),
where u = (u1 , u2 , . . . , uk ).
Exercise 8.2 The smallest Hamming code is a binary code [7, 4, 3], which is a
single error correcting code. It has the following generating matrix (in standard
form): G = [I4 |A] where


1 1 1
 0 1 0 

A=
 1 0 1 .
1 1 0
Find a parity-check matrix for it.
8.2
Codes from Designs
The code CF of the design D over the finite field F is the space spanned by
the incidence vectors of the blocks over F . If we take F to be a prime field
Fp = GF (p), in which case we write also Cp for CF , and refer to the dimension
of Cp as the p-rank of D. If the point set of D is denoted by P and the block set
by B, and if Q is any
subset of P, then we will denote the incidence vector of Q
by v Q . Thus CF = v B | B ∈ B , and is a subspace of F P , the full vector space
of functions from P to F .
Example 8.1 Let D be the 2-design representing the Fano plane. Then using
the Exercise 7.2 we can easily see that
i. If F is a field of characteristic p with p ∈
/ {2, 3}, then CF (D) = F 7 with
∼
Aut(C) = S7 .
9
METHOD 1
53
ii. C2 (D) = [7, 4, 3]2 (the smallest hamming code) with Aut(C) ∼
= P GL(3, 2) ∼
=
P SL(2, 7) ∼
P
SL(3,
2),
the
simple
group
of
order
168.
=
iii. C3 (D) =<  >⊥ = [7, 6, 2]3 , with Aut(C) ∼
= S7 .
Terminology for graphs is standard: our graphs are undirected, the valency of
a vertex is the number of edges containing the vertex. A graph is regular if all
the vertices have the same valence, and a regular graph is strongly regular of
type (n, k, λ, µ) if it has n vertices, valence k, and if any two adjacent vertices are
together adjacent to λ vertices, while any two non-adjacent vertices are together
adjacent to µ vertices.
9
Method 1
Construction of 1-Designs and Codes from Maximal Subgroups: In this
section we consider primitive representations of a finite group G. Let G be a
finite primitive permutation group acting on the set Ω of size n. We can consider
the action of G on Ω × Ω given by (α, β)g = (αg , β g ) for all α, β ∈ Ω and all
¯ is an orbital, then
g ∈ G. An orbit of G on Ω × Ω is called an orbital. If ∆
∗
¯
¯
∆ = {(α, β) : (β, α) ∈ ∆} is also an orbital of G on Ω × Ω, which is called the
¯ We say that ∆
¯ is self-paired if ∆
¯ =∆
¯ ∗.
paired orbital of ∆.
Now Let α ∈ Ω, and let ∆ 6= {α} be an orbit of the stabilizer M = Gα of
¯ given by ∆
¯ = {(α, δ)g : δ ∈ ∆, g ∈ G} is an
α. It is not difficult to see that ∆
¯ is a self paired orbital. Also
orbital. We say that ∆ is self-paired if and only if ∆
note that the primitivity of G on Ω implies that M is a maximal subgroup of G.
If M = Gα has only three orbits {α}, ∆ and ∆0 on Ω, then we say that G is
a rank-3 permutation group.
Our construction for the symmetric 1-designs is based on the following results,
mainly Theorem 9.1 below, which is the Proposition 1 of [18] with its corrected
version in [19]:
Theorem 9.1 Let G be a finite primitive permutation group acting on the set Ω
of size n. Let α ∈ Ω, and let ∆ 6= {α} be an orbit of the stabilizer Gα of α. If
B = {∆g : g ∈ G}
and, given δ ∈ ∆,
E = {{α, δ}g : g ∈ G},
then D = (Ω, B) forms a 1-(n, |∆|, |∆|) design with n blocks. Further, if ∆ is a
self-paired orbit of Gα , then Γ = (Ω, E) is a regular connected graph of valency |∆|,
D is self-dual, and G acts as an automorphism group on each of these structures,
primitive on vertices of the graph, and on points and blocks of the design.
Proof: We have |G| = |∆G ||G∆ |, and clearly G∆ ⊇ Gα . Since G is primitive on
Ω, Gα is maximal in G, and thus G∆ = Gα , and |∆G | = |B| = n. This proves
that we have a 1-(n, |∆|, |∆|) design.
Since ∆ is self-paired, Γ is a graph rather than only a digraph. In Γ we notice
that the vertices adjacent to α are the vertices in ∆. Now as we orbit these pairs
9
METHOD 1
54
under G, we get the nk ordered pairs, and thus nk/2 edges, where k = ∆. Since
the graph has G acting, it is clearly regular, and thus the valency is k as required,
i.e. the only vertices adjacent to α are those in the orbit ∆. The graph must be
connected, as a maximal connected component will form a block of imprimitivity,
contradicting the group’s primitive action.
Now notice that an adjacency matrix for the graph is simply an incidence
matrix for the 1-design, so that the 1-design is necessarily self-dual. This proves
all our assertions. Note that if we form any union of orbits of the stabilizer of a point, including
the orbit consisting of the single point, and orbit this under the full group, we will
still get a self-dual symmetric 1-design with the group operating. Thus the orbits
of the stabilizer can be regarded as “building blocks”. Since the complementary
design (i.e. taking the complements of the blocks to be the new blocks) will have
exactly the same properties, we will assume that our block size is at most v/2.
In fact this will give us all possible designs on which the group acts primitively
on points and blocks:
Lemma 9.2 If the group G acts primitively on the points and the blocks of a
symmetric 1-design D, then the design can be obtained by orbiting a union of
orbits of a point-stabilizer, as described in Theorem 9.1.
Proof: Suppose that G acts primitively on points and blocks of the 1-(v, k, k)
design D. Let B be the block set of D; then if B is any block of D, B = B G . Thus
|G| = |B||GB |, and since G is primitive, GB is maximal and thus GB = Gα for
some point. Thus Gα fixes B, so this must be a union of orbits of Gα . Lemma 9.3 If G is a primitive simple group acting on Ω, then for any α ∈ Ω,
the point stabilizer Gα has only one orbit of length 1.
Proof: Suppose that Gα fixes also β. Then Gα = Gβ . Since G is transitive,
there exists g ∈ G such that αg = β. Then (Gα )g = Gαg = Gβ = Gα , and thus
g ∈ NG (Gα ) = N , the normalizer of Gα in G. Since Gα is maximal in G, we have
N = G or N = Gα . But G is simple, so we must have N = Gα , so that g ∈ Gα
and so β = α. We have considered various finite simple groups, for example J1 ; J2 ; M c L;
P Sp2m (q), where q is a power of an odd prime, and m ≥ 2; Co2 ; HS and Ru.
For each group, using Magma [4], we construct designs and graphs that have the
group acting primitively on points as automorphism group, and, for a selection
of small primes, codes over that prime field derived from the designs or graphs
that also have the group acting as automorphism group. For each code, the code
automorphism group at least contains the associated group G.
To aid in the classification, if possible, the dimension of the hull of the design
for each of these primes were found. Then we took a closer look at some of the
more interesting codes that arose, asking what the basic coding properties were,
and if the full automorphism group could be established.
It is well known, and easy to see, that if the group is rank-3, then the graph
formed as described in Theorem 9.1 will be strongly regular. In case the group
9
METHOD 1
55
is not of rank 3, this might still happen, and we examined this question also for
some of the groups we studied.
A sample of our results for example for J1 and J2 is given below, but the full
set can be obtained at Jenny Key’s web site under the file “Janko groups and
designs”:
http://www.ces.clemson.edu/~keyj
Clearly the automorphism group of any of the codes will contain the automorphism group of the design from which it is formed. We looked at some of
the codes that were computationally feasible to find out if the groups J1 and J¯2
formed the full automorphism group in any of the cases when the code was not
the full vector space. We first mention the following lemma:
Lemma 9.4 Let C be the linear code of length n of an incidence structure I over
a field F. Then the automorphism group of C is the full symmetric group if and
only if C = F n or C = F ⊥ .
Proof: Suppose Aut(C) is Sn . C is spanned by the incidence vectors of the blocks
of I; let B be such a block and suppose it has k points, and so it gives a vector of
weight k in C. Clearly C contains the incidence vector of any set of k points, and
thus, by taking the difference of two such vectors that differ in just two places, we
see that C contains all the vectors of weight 2 having as non-zero entries 1 and
−1. Thus C = F ⊥ or F n . The converse is clear.
Huffman [15] has more on codes and groups, and in particular, on the possibility of the use of permutation decoding for codes with large groups acting. See
also Knapp and Schmid [26] for more on codes with prescribed groups acting.
Most of the codes we looked at were too large to find the automorphism group,
but we did find some of, through computation with Magma. Note that we could
in some cases look for the full group of the hull, and from that deduce the group
of the code, since Aut(C) = Aut(C ⊥ ) ⊆ Aut(C ∩ C ⊥ ).
9.1
J1 , J2 and Co2
In this subsection we give a brief discussion on the application of Method 1 (discussed in Section 9) to the sporadic simple groups J1 , J2 and Co2 . For full details
the readers are referred to [18], [19], [20] and [32].
9.1.1
Computations for J1 and J2
The first Janko sporadic simple group J1 has order 175560 = 23 ×3×5×7×11×19
and it has seven distinct primitive representations, of degree 266, 1045, 1463, 1540,
1596, 2926, and 4180, respectively (see Table 1 and [5, 10]). For each of the seven
primitive representations, using Magma, we constructed the permutation group
and formed the orbits of the stabilizer of a point. For each of the non-trivial
orbits, we formed the symmetric 1-design as described in Theorem 9.1. We took
set of the {2, 3, 5, 7, 11} of primes and found the dimension of the code and its hull
for each of these primes. Note also that since 19 is a divisor of the order of J1 , in
some of the smaller cases it is worthwhile also to look at codes over the field of
9
METHOD 1
56
order 19. We also found the automorphism group of each design, which will be the
same as the automorphism group of the regular graph. Where computationally
possible we also found the automorphism group of the code.
Conclusions from our results are summarized below. In brief, we found that
there are 245 designs formed in this manner from single orbits and that none of
them is isomorphic to any other of the designs in this set. In every case the full
automorphism group of the design or graph is J1 .
No.
Max[1]
Max[2]
Max[3]
Max[4]
Max[5]
Max[6]
Max[7]
Order
660
168
120
114
110
60
42
Index
266
1045
1463
1540
1596
2926
4180
Structure
P SL(2, 11)
23 :7:3
2 × A5
19:6
11:10
D6 × D10
7:6
Table 1: Maximal subgroups of J1
In Table 2, the first column gives the degree, the second the number of orbits,
and the remaining columns give the length of the orbits of length greater than
1, with the number of that length in parenthesis behind the length in case there
is more than one of that length. The pairs that had the same code dimensions
occurred as follows: for degrees 266, 1045 and 1596, there were no such pairs; for
degree 1463 there were two pairs, both for orbit size 60; for degree 1540, there
were two pairs, for orbit size 57 and 114 respectively; for degree 2926 there was
one pair for orbit size 60; for degree 4180 there were 12 pairs, for orbit size 42.
In summary then, we have the following:
Proposition 9.5 If G is the first Janko group J1 , there are precisely 245 nonisomorphic self-dual 1-designs obtained by taking all the images under G of the
non-trivial orbits of the point stabilizer in any of G’s primitive representations,
and on which G acts primitively on points and blocks. In each case the full automorphism group is J1 . Every primitive action on symmetric 1-designs can be
Degree
266
1045
1463
1540
1596
2926
4180
#
5
11
22
21
19
67
107
length
132
168(5)
120(7)
114(9)
110(13)
60(34)
42(95)
110
56(3)
60(9)
57(6)
55(2)
30(27)
21(6)
12
28
20(2)
38(4)
22(2)
15(5)
14(4)
11
8
15(2)
19
11
7
Table 2: Orbits of a point-stabilizer of J1
12
9
METHOD 1
57
obtained by taking the union of such orbits and orbiting under G.
We tested the graphs for strong regularity in the cases of the smaller degree,
and did not find any that were strongly regular. We also found the designs and
their codes for some of the unions of orbits in some cases. We found that some
of the codes were the same for some primes, but not for all.
The second Janko sporadic simple group J2 has order has order 604800 =
27 × 33 × 52 × 7, and it has nine primitive permutation representations (see Table
3), but we did not compute with the largest degree. Thus our results cover only
the first eight. Our results for J2 are different from those for J1 , due to the
existence of an outer automorphism. The main difference is that usually the full
automorphism group is J¯2 , and that in the cases where it was only J2 , there would
be another orbit of that length that would give an isomorphic design, and which,
if the two orbits were joined, would give a design of double the block size and
automorphism group J¯2 . A similar conclusion held if some union of orbits was
taken as a base block.
No.
Max[1]
Max[2]
Max[3]
Max[4]
Max[5]
Max[6]
Max[7]
Max[8]
Max[9]
Order
6048
2160
1920
1152
720
600
336
300
60
Index
100
280
315
525
840
1008
1800
2016
10080
Structure
P SU (3, 3)
3. P GL(2, 9)
21+4 :A5
22+4 :(3 × S3 )
A4 × A5
A5 × D10
P SL(2, 7):2
52 :D12
A5
Table 3: Maximal subgroups of J2
From these eight primitive representations, we obtained in all 51 non-isomorphic
symmetric designs on which J2 acts primitively. Table 4 gives the same information for J2 that Table 2 gives for J1 . The automorphism group of the design in
Degree
100
280
315
525
840
1008
1800
2016
#
3
4
6
6
7
11
18
18
length
63
135
160
192(2)
360
300
336
300(2)
36
108
80
96
240
150(2)
168(6)
150(6)
36
32(2)
32
180
100(2)
84(3)
75(5)
10
12
24
60(2)
42(3)
50(2)
20
50
28
25
15
25
21
15
12
14(2)
Table 4: Orbits of a point-stabilizer of J2 (of degree ≤ 2016)
9
METHOD 1
58
each case was J2 or J¯2 . Where J2 was the full group, there is another copy of
the design for another orbit of the same length. This occurred in the following
cases: degree 315, orbit length 32; degree 1008, orbit lengths 60, 100 and 150;
degree 1800, orbit lengths 42, 42, 84 and 168; degree 2016, orbit lengths 50, 75,
75, 150, 150, and 300. We note again that the p-ranks of the design and their hulls
gave an initial indication of possible isomorphisms and clear non-isomorphisms,
so that only the few mentioned needed be tested. This reduced the computations
tremendously.
We also found three strongly regular graphs (all of which are known: see
Brouwer [7]): that of degree 100 from the rank-3 action, of course, and two more
of degree 280 from the orbits of length 135 and 36, giving strongly regular graphs
with parameters (280,135,70,60) and (280,36,8,4) respectively. The full automorphism group is J¯2 in each case. We have not checked all the other representations
but note that this is the only one with point stabilizer having exactly four orbits.
Note that Bagchi [3] found a strongly regular graph with J2 acting.
In each of the following we consider the primitive action of J2 on a design
formed as described in Method 1 from an orbit or a union of orbits, and the codes
are the codes of the associated 1-design.
1. For J2 of degree 100, J¯2 is the full automorphism group of the design with
parameters 1-(100, 36, 36), and it is the automorphism group of the selforthogonal doubly-even [100, 36, 16]2 binary code of this design.
2. For J2 of degree 280, J¯2 is the full automorphism group of the design with
parameters 1-(280, 108, 108), and it is the automorphism group of the selforthogonal doubly-even [280, 14, 108]2 binary code of this design. The weight
distribution of this code is
<0, 1>, <108, 280>, <128, 1575>, <136, 2520>, <140, 7632>,
<144, 2520>, <152, 1575>, <172, 280>, <280, 1>
Thus the words of minimum weight (i.e. 108) are the incidence vectors of
the design.
3. For J2 of degree 315, J¯2 is the full automorphism group of the design with
parameters 1-(315, 64, 64) (by taking the union of the two orbits of length
32), and it is the automorphism group of the self orthogonal doubly-even
[315, 28, 64]2 binary code of this design. The weight distribution of the code
is as follows:
<0, 1>,<64, 315>,<96, 6300>,<104, 25200>,<112, 53280>, <120,
242760>,<124, 201600>,<128, 875700>,<132, 1733760>, <136,
4158000>,<140, 5973120>,<144, 12626880>,<148, 24232320>, <152,
35151480>,<156, 44392320>,<160, 53040582>, <164, 41731200>,<168,
28065120>,<172, 13023360>,<176, 2129400>, <180, 685440>,<184,
75600>,<192, 10710>,<200, 1008>
Thus the words of minimum weight (i.e. 64) are the incidence vectors of the
blocks of the design.
9
METHOD 1
59
Furthermore, the designs from the two orbits of length 32 in this case, i.e.
1-(315, 32, 32) designs, each have J2 as their automorphism group. Their binary codes are equal, and are [315, 188]2 codes, with hull the 28-dimensional
code described above. The automorphism group of this 188-dimensional
code is again J¯2 . The minimum weight is at most 32. This is also the
binary code of the design from the orbit of length 160.
4. For J2 of degree 315, J¯2 is the full automorphism group of the design
with parameters 1-(315, 160, 160) and it is the automorphism group of the
[315, 265]5 5-ary code of this design. This code is also the 5-ary code of the
design obtained from the orbit of length 10, and from that of the orbit of
length 80, so we can deduce that the minimum weight is at most 10. The
hull is a [315, 15, 155]5 code and again with J¯2 as full automorphism group.
5. For J2 of degree 315, J¯2 is the full automorphism group of the design with
parameters 1-(315, 80, 80) from the orbit of length 80, and it is the automorphism group of the self-orthogonal doubly-even [315, 36, 80]2 binary code of
this design. The minimum words of this code are precisely the 315 incidence
vectors of the blocks of the design.
In [20] we used the construction described in Method 1 to obtain all irreducible
modules of J1 (as codes) over the prime fields F2 , F3 , F5 . We also showed that
most of those of J2 can be represented in this way as the code, the dual code or
the hull of the code of a design, or of codimension 1 in one of these. For J2 , if no
such code was found for a particular irreducible module, then we checked that it
could not be so represented for the relevant degrees of the primitive permutation
representations up to and including 1008. In summary, we obtained:
Proposition 9.6 Using the construction described in Method 1 above (see Theorem 9.1 and Lemma 9.2), taking unions of orbits, the following constructions of
the irreducible modules of the Janko groups J1 and J2 as the code, the dual code
or the hull of the code of a design, or of codimension 1 in one of these, over Fp
where p = 2, 3, 5, were found to be possible:
1. J1 : all the seven irreducible modules for p = 2, 3, 5;
2. J2 : all for p = 2 apart from dimensions 12, 128; all for p = 3 apart from dimensions 26, 42, 114, 378; all for p = 5 apart from dimensions 21, 70, 189, 300.
For these exclusions, none exist of degree ≤ 1008.
Note: 1. We do not claim that we have all the constructions of the modular
representations as codes; we were seeking mainly existence.
We give below three self-orthogonal binary codes of dimension 20 invariant
under J1 of lengths 1045, 1463, and 1540. These are irreducible by [16] or Magma
data. In all cases the Magma simgps library is used for J1 and J2 .
1. J1 of Degree 1045
[1045, 20, 456]2 code; dual code: [1045, 1025, 4]2
9
METHOD 1
60
\\Orbit lengths of stabilizer of a point:
[ 1, 8, 28, 56, 56, 56, 168, 168, 168, 168, 168 ];
\\Orbits chosen: ##1,3,5,10,11
\\Defining block is the union of these, length 421
1-(1045, 421, 421) Design with 1045 blocks
\\C is the code of the design, of dimension 21
\\The 20-dimensional code is Ch:= C meet Dual(C) =Hull(C)
> WeightDistribution(Ch);
[ <0, 1>, <456, 3080>, <488, 29260>, <496, 87780>, <504, 87780>,
<512, 36575>, <520, 299706>, <528, 234080>, <536, 175560>, <544,
58520>, <552, 14630>, <560, 19019>, <608, 1540>, <624, 1045> ].
Those of weight 456, 504, 544, 552, 624, 608 are single orbits; the
others split.
>WeightDistribution(C);
[ <0, 1>, <421,1405>, <437, 1540>, <456, 3080>, <485,19019>,
<488, 29260>, <493, 14630>, < 496, 87780>, <501, 58520>,
<504, 87780>, <509, 175560>, <512, 36575>, <517, 234080>, <520,
299706>, <525, 299706>, <528, 234080>, <533, 36575>, <536, 175560>,
<541, 87780>, <544, 58520>, <549, 87780>, <552, 14630>, <557,29260>,
<560, 19019>, <589, 3080>, <608, 1540>, <624, 1045>, <1045, 1> ].
2. J1 of Degree 1463
[1463, 20, 608]2 code; dual code: [1463, 1443, 3]2
\\Orbit lengths of stabilizer of a point:
[ 1, 12, 15, 15, 20, 20, 60, 60, 60, 60, 60, 60, 60, 60, 60, 120,
120, 120, 120, 120, 120, 120 ]
\\Orbits chosen ##18,21
\\Defining block is union of these, of length 240
1-(1463, 240, 240) Design with 1463 blocks
\\C is the code of the design, of dimension 492
\\The 20-dimensional code is Ch:= C meet Dual(C) =Hull(C)
WD(Ch);
[ <0, 1>, <608, 1540>, <632, 2926>, <640, 7315>, <688, 29260>, <696,
29260>, <712, 87780>, <720, 89243>, <728, 311410>, <736, 87780>,
<744, 175560>, <752, 222376>, <760, 3080>, <784, 1045> ]
3. J1 of Degree 1540
[1540, 20, 640]2 code; dual code: [1540, 1520, 4]2
\\Orbit lengths of stabilizer of a point:
[ 1, 19, 38, 38, 38, 38, 57, 57, 57, 57, 57, 57, 114, 114, 114, 114,
114, 114, 114, 114, 114 ]
\\Orbits chosen ##7,13
\\Defining block is the union of these, length 171
1-(1540, 171, 171) Design with 1540 blocks
\\C is the code of the design, of dimension 592
\\The code of dimension 20 is Ch:=C meet Dual(C)
WD(Ch); [ <0, 1>, <640, 1463>, <728, 33440>, <736, 58520>, <760,
311696>, <768, 358435>, <792, 175560>, <800, 105336>, <856, 3080>,
<896, 1045> ]
9
METHOD 1
61
We now look at the smallest representations for J2 . We have not been able
to find any of dimension 12, and none can exist for degree ≤ 1008, as we have
verified computationally by examining the permutation modules. We give below
four representations of J2 acting on self-orthogonal binary codes of small degree
that are irreducible or indecomposable codes over J2 . The full automorphism
group of each of these codes is J¯2 .
1. J2 of Degree 100, dimension 36
[100, 36, 16]2 code; dual code: [100, 64, 8]2
\\Orbit lengths of stabilizer of a point:
[1, 36, 63] 1-(100, 36, 36) Design with 100 blocks
\\ Orbit #2 gave a block of the design
[ <0, 1>, <16, 1575>, <24, 105000>, <28, 1213400>, <32, 29115450>,
<36, 429677200>, <40, 2994639480>, <44, 10672216200>, <48,
20240374350>, <52, 20217640800>, <56, 10675819800>, <60,
3004193640>, <64, 422248725>, <68, 30819600>, <72, 1398600>, <76,
12600>, <80, 315> ]
This code C = C36 of dimension 36 is irreducible, by Magma. The dual code
C64 = C ⊥ has an invariant subcode C63 of dimension 63 that is spanned
by the weight-8 vectors and that contains  and C36 . All these codes are
indecomposable, by Magma. The full automorphism group of this code is
J¯2 .
2. J2 of Degree 280, dimension 13
[280, 13, 128]2 code; dual code: [280, 267, 4]2
\\Orbit lengths of stabilizer of a point:
[1, 36, 108, 135]
\\Orbit #3 gave a block of the design
1-(280,108,108) Design with 280 blocks
\\Weight distribution of its 14-dimensional binary code
[ <0, 1>, <108, 280>, <128, 1575>, <136, 2520>, <140, 7632>, <144,
2520>, <152, 1575>, <172, 280>, <280, 1> ] Dual code: [280,266,4]
\\Weight distribution of reducible but indecomposable 13-dimensional code
[ <0, 1>, <128, 1575>, <136, 2520>, <144, 2520>, <152, 1575>, <280,
1> ]
This code has the invariant subcode of dimension 1 generated by the allone vector, so it is reducible. However, we checked the orbits of all the
other words and found that there are no other invariant subcodes. It is thus
indecomposable. The full automorphism group of these codes is J¯2 .
3. J2 of Degree 315, dimension 28
[315, 28, 64]2 code; dual code: [315, 287, 3]2
\\Orbit lengths of stabilizer of a point:
[ 1, 10, 32, 32, 80, 160 ]
9
METHOD 1
62
\\Orbits ## 3 and 4 chosen
1-(315, 64, 64) Design with 315 blocks
\\Weight distribution of its 28-dimensional binary code
[ <0, 1>, <64, 315>, <96, 6300>, <104, 25200>, <112, 53280>, <120,
242760>, <124, 201600>, <128, 875700>, <132, 1733760>, <136,
4158000>, <140, 5973120>, <144, 12626880>, <148, 24232320>, <152,
35151480>, <156, 44392320>, <160, 53040582>, <164, 41731200>, <168,
28065120>, <172, 13023360>, <176, 2129400>, <180, 685440>, <184,
75600>, <192, 10710>, <200, 1008> ]
The code is an irreducible module over J2 , by Magma. The full automorphism group of this code is J¯2 .
4. J2 of Degree 315, dimension 36
[315, 36, 80]2 code; dual code: [315, 279, 5]2
\\Orbit lengths of stabilizer of a point:
[ 1, 10, 32, 32, 80, 160 ]
\\chose the orbit of length 80
1-(315, 80, 80) Design with 315 blocks 36 =Dim(C) dim hull 36
//Weight distribution of the 36-dimensional code
[ <0, 1>, <80, 315>, <84, 1800>, <96, 9450>, <100, 50400>, <108,
126000>, <112, 84150>, <116, 466200>, <120, 4798920>, <124,
10987200>, <128, 54432000>, <132, 180736920>, <136, 606475800>,
<140, 1792977480>, <144, 3988438335>, <148, 6923044800>, <152,
10151396640>, <156, 12278475300>, <160, 11844516600>, <164,
9314451720>, <168, 6136980600>, <172, 3360636720>, <176,
1436425200>, <180, 459183200>, <184, 132924960>, <188, 32715900>,
<192, 7006125>, <196, 1800000>, <200, 126000>, <204, 113400>, <208,
75600>, <216, 12600>, <220, 6300>, <252, 100> ]
The code is an irreducible module over J2 , by Magma. The full automorphism group of this code is J¯2 .
For F one of the fields Fp for p = 2, 3, 5 and n the degree of the permutation
representation, in [20] we demonstrated some cases where the full space F n can be
completely decomposed into G-modules, where G = J1 , J2 , using codes obtained
by our construction. In all cases Cm denotes an indecomposable linear code of
dimension m over the relevant field and group. If the codes were irreducible they
were obtained according to our method and were listed in [20]. For example
• For J1 of degree 1045 over F2 , the full space can be completely decomposed
into J1 -modules, that is:
F1045
= C76 ⊕ C112 ⊕ C360 ⊕ C496 ⊕ F2 ,
2
where all but C496 are irreducible. C496 has composition factors of dimensions 20, 112, 1, 76, 20, 1, 112, 20, 1, 1, 112, 20. Also
S = Socle(F1045
) = F2  ⊕ C20 ⊕ C76 ⊕ C112 ⊕ C360 ,
2
with dim(S) = 569.
9
METHOD 1
63
• For J2 of degree 315 over F2 we have:
F2315 = C160 ⊕ C154 ⊕ F2 ,
⊥
where C160 is irreducible and C154 ⊕ F2  = C160
is the binary code of the
1-(315, 33, 33) design from orbits #1 and #4. (Note that F100
and F280
are
2
2
indecomposable as J2 modules.)
• For J2 of degree 100 over F3 we have:
F3100 = C36 ⊕ C63 ⊕ F3 .
• For J2 of degree 280 over F3 we have:
F3280 = C63 ⊕ C216 ⊕ F3 ,
where C216 is the code of the 1-(280, 135, 135) design obtained from the orbit
# 4.
• for J2 of degree 525 over F5 we have:
F5525 = C175 ⊕ C100 ⊕ C250 ,
where C175 is irreducible and C100 is the dual of the code C of the 1(525, 140, 140) design obtained from the orbits #2, #3 , #4, and C250 =
⊥
C ∩ C175
.
9.1.2
The Conway group Co2
The Leech lattice is a certain 24-dimensional Z submodule of the Euclidean space
R24 whose automorphism group is the double cover 2. Co1 of the Conway group
Co1 . The Conway groups Co2 and Co3 are stabilizers of sublattices of the Leech
lattice.
The subgroup structure of Co2 is discussed in Wilson [40] and [39] using the
following information. The group Co2 admits a 23-dimensional indecomposable
representation over GF (2) obtained from the 24-dimensional Leech lattice by
reducing modulo 2 and factoring out a fixed vector. The action of Co2 on the
vectors of this 23-dimensional indecomposable GF (2) module (say M ) produces
eight orbits, with stabilizers isomorphic to Co2 , U6 (2):2, 210 :M22 :2, M c L, HS:2,
U4 (3).D8 , 21+8
+ :S8 and M23 , respectively. The 23-dimensional indecomposable
GF (2) module M contains an irreducible GF (2)-submodule N of dimension 22.
We use TABLE III(a) given by Wilson in [39] to produce Table 5, which gives
the orbit lengths and stabilizers for the actions of Co2 on M and N respectively.
On the other hand, reduction modulo 2 of the 23-dimensional ordinary irreducible representation results in a decomposable 23-dimensional GF (2) representation. In [40] Wilson showed that Co2 has exactly eleven conjugacy classes of
maximal subgroups. One of these subgroups is the group U6 (2):2 of index 2300.
In Proposition 9.7, using this maximal subgroup, we construct the decomposable
23-dimensional GF (2)-representation as the binary code C892 of dimension 23 invariant under the action of Co2 . The action of Co2 on C892 produces 12 orbits
9
METHOD 1
M -Stabilizer
Co2
U6 (2) : 2
64
M -Orbit length
1
2300
c
M L
47104
10
46575
2 :M22 :2
HS:2
476928
U4 (3).D8
1619200
M23
4147200
21+8
+ :S8
2049300
N -Stabilizer
N -Orbit length
Co2
1
U6 (2) : 2
2300
210 :M22 :2
46575
HS:2
476928
U4 (3).D8
1619200
21+8
+ :S8
2049300
Table 5: Action of Co2 on M and N
with stabilizers isomorphic to Co2 (2 copies), U6 (2):2 (2 copies), 210 :M22 :2 (2
: S8 (2 copies) respectively.
copies), HS:2 (2 copies), U4 (3).D8 (2 copies), 21+8
+
Furthermore, C892 contains a binary code C1408 of dimension 22 invariant and
irreducible under the action of Co2 . Notice that the 2-modular character table
of Co2 is completely known (see [36]) and follows from it that the irreducible
22-dimensional GF (2) representation is unique and 22 is the smallest dimension
for any non-trivial irreducible GF (2) module.
Here we examine some designs Di and associated binary codes Ci constructed
from a primitive permutation representation of degree 2300 of the sporadic simple
group Co2 . For the full detail the readers are encouraged to see [32].
We used Method 1 and constructed self-dual symmetric 1-designs Di and
binary codes Ci , where i is an element of the set {891, 892, 1408, 1409, 2299}, from
the rank-3 primitive permutation representation of degree 2300 of the sporadic
simple group Co2 of Conway. The stabilizer of a point α in this representation
is a maximal subgroup isomorphic to U6 (2):2, producing orbits {α}, ∆1 , ∆2 of
lengths 1, 891 and 1408 respectively.
The self-dual symmetric 1-designs Di are constructed from the sets ∆1 , {α} ∪
∆1 , ∆2 , {α} ∪ ∆2 , and ∆1 ∪ ∆2 , respectively. We let Ω = {α} ∪ ∆1 ∪ ∆2 .
We proved the following result:
Proposition 9.7 Let G be the Conway group Co2 and Di and Ci where i is in
the set {891, 892, 1408, 1409, 2299} be the designs and binary codes constructed
from the primitive rank-3 permutation action of G on the cosets of U6 (2):2. Then
the following holds:
(i) Aut(D891 ) = Aut(D892 ) = Aut(D1408 ) =
Aut(D1409 ) = Aut(C892 ) = Aut(C1408 ) = Co2 .
(ii) dim(C892 ) = 23, dim(C1408 ) = 22,
C892 ⊃ C1408 and Co2 acts irreducibly on C1408 .
10
METHOD 2
65
(iii) C891 = C1409 = C2299 = V2300 (GF (2)).
(iv) Aut(D2299 ) = Aut(C891 ) = Aut(C1049 ) =
Aut(C2299 ) = S2300 .
The proof of the proposition follows from a series of lemmas. In fact we showed
that the codes C892 and C1408 are of types [2300, 23, 892]2 and [2300, 22, 1024]2
respectively. Furthermore
C892 = hC1408 , i = C1408 ∪ {w +  : w ∈ C1408 }
= C1408 ⊕ hi,
where  denotes the all-one vector. Let Wl denote the set of all codewords of C892
of weight l and let Al be the size of Wl . Then clearly Wl + {} = W2300−l ⊂ C892
and |Wl | = Al = |W2300−l | = A2300−l . We found the weight distribution of
C892 and then the weight distribution of C1408 follows. We also determined the
structures of the stabilizers (Co2 )wl , for all nonzero weight l, where wl ∈ C1408 is
a codeword of weight l. The structures of the stabilizers (Co2 )wl for C892 follows
clearly from those of C1408 .
We also showed that the code C1408 is the 22 dimensional irreducible representation of Co2 over GF (2) contained in the 23-dimensional decomposable C892 .
It is also contained in the 23-dimensional indecomposable representation of Co2
over GF (2) discussed in ATLAS [5] and Wilson [39].
We should also mention that computation with Magma shows the codes over
some other primes, in particular, p = 3 are of some interest. In a separate paper
we plan to deal with the ternary codes invariant under Co2 [35].
10
Method 2
Construction of 1-Designs and Codes from Maximal Subgroups and
Conjugacy Classes of Elements: In this section we assume G is a finite simple
group, M is a maximal subgroup of G, nX is a conjugacy class of elements of
order n in G and g ∈ nX. Thus Cg = [g] = nX and |nX| = |G : CG (g)|.
As in Section 6 let χM = χ(G|M ) be the permutation character afforded by
the action of G on Ω, the set of all conjugates of M in G. Clearly if g is not
conjugate to any element in M , then χM (g) = 0.
The construction of our 1-designs is based on the following theorem.
Theorem 10.1 Let G be a finite simple group, M a maximal subgroup of G and
nX a conjugacy class of elements of order n in G such that M ∩ nX 6= ∅. Let
B = {(M ∩nX)y |y ∈ G} and P = nX. Then we have a 1−(|nX|, |M ∩nX|, χM (g))
design D, where g ∈ nX. The group G acts as an automorphism group on D,
primitive on blocks and transitive (not necessarily primitive) on points of D.
Proof: First note that
B = {M y ∩ nX|y ∈ G}.
We claim that M y ∩ nX = M ∩ nX if and only if y ∈ M or nX = {1G }. Clearly
if y ∈ M or nX = {1G }, then M y ∩ nX = M ∩ nX. Conversely suppose there
10
METHOD 2
66
exits y ∈
/ M such that M y ∩ nX = M ∩ nX. Then maximality of M in G implies
that G =< M, y > and hence M z ∩ nX = M ∩ nX for all z ∈ G. We can deduce
that nX ⊆ M and hence < nX >≤ M. Since < nX > is a normal subgroup of G
and G is simple, we must have < nX >= {1G }. Note that maximality of M and
the fact < nX >≤ M , excludes the case < nX >= G.
From above we deduce that
b = |B| = |Ω| = [G : M ].
If B ∈ B, then
k = |B| = |M ∩ nX| =
k
X
|[xi ]M | = |M |
i=1
k
X
i=1
1
,
|CM (xi )|
where x1 , x2 , ..., xk are the representatives of the conjugacy classes of M that fuse
to g.
Let v = |P| = |nX| = [G : CG (g)]. Form the design D = (P, B, I), with point
set P, block set B and incidence I given by xIB if and only if x ∈ B. Since the
number of blocks containing an element x in P is λ = χM (x) = χM (g), we have
produced a 1 − (v, k, λ) design D, where v = |nX|, k = |M ∩ nX| and λ = χm (g).
The action of G on blocks arises from the action of G on Ω and hence the
maximality of M in G implies the primitivity. The action of G on nX, that is on
points, is equivalent to the action of G on the cosets of CG (g). So the action on
points is primitive if and only if CG (g) is a maximal subgroup of G. Remark 10.1 Since in a 1 − (v, k, λ) design D we have kb = λv, we deduce that
k = |M ∩ nX| =
χM (g) × |nX|
.
[G : M ]
Also note that D̃, the complement of D, is 1 − (v, v − k, λ̃) design, where λ̃ =
λ × v−k
k .
Remark 10.2 If λ = 1, then D is a 1 − (|nX|, k, 1) design. Since nX is the
disjoint union of b blocks each of size k, we have Aut(D) = Sk o Sb = (Sk )b : Sb .
Clearly In this case for all p, we have C = Cp (D) = [|nX|, b, k]p , with Aut(C) =
Aut(D).
Remark 10.3 The designs D constructed by using Theorem 12 are not symmetric
in general. In fact D is symmetric if and only if
b = |B| = v = |P| ⇔ [G : M ] = |nX| ⇔ [G : M ] = [G : CG (g)] ⇔ |M | = |CG (g)|.
10.1
Some 1-designs and Codes from A7
A7 has five conjugacy classes of maximal subgroups, which are listed in Table 6.
It has also 9 conjugacy classes of elements, some of which are listed in Table 7.
We apply the Theorem 10.1 to the above maximal subgroups and few conjugacy classes of elements of A7 to construct several non-symmetric 1- designs. The
corresponding binary codes are also constructed.
10
METHOD 2
67
No.
Max[1]
Max[2]
Max[3]
Max[4]
Max[5]
Structure
A6
P SL2 (7)
P SL2 (7)
S5
(A4 × 3):2
Index
7
15
15
21
35
Order
360
168
168
120
72
Table 6: Maximal subgroups of A7
nX
2A
3A
3B
|nX|
105
70
280
CG (g)
D8 : 3
A4 × 3 ∼
= (22 × 3): 3
3×3
Maximal Centralizer
No
No
No
Table 7: Some of the conjugacy classes of A7
10.1.1
G = A7 , M = A6 and nX = 3A
Let G = A7 , M = A6 and nX = 3A. Then
b = [G : M ] = 7, v = |3A| = 70, k = |M ∩ 3A| = 40.
Also using the character table of A7 , we have χM = χ1 + χ2 = 1a + 6a and hence
χM (g) = 1+3 = 4 = λ, where g ∈ 3A. We produce a non-symmetric 1−(70, 40, 4)
design D. A7 acts primitively on the 7 blocks. Since CA7 (g) = A4 × 3 is not
maximal in A7 (sits in the maximal subgroup (A4 × 3):2 with index two), A7
acts imprimitively on the 70 points. The complement of D, D̃, is a 1 − (70, 30, 3)
design.
Computations with MAGMA [4] shows that the full automorphism group of
D is
Aut(D) ∼
= 235 :S7 ∼
= 25 o S7 ,
with |Aut(D)| = 239 .32 .5.7. Construction using MAGMA shows that the binary
code C of this design is a [70, 6, 32]2 code. The code C is self-orthogonal with the
weight distribution
< 0, 1 >, < 32, 35 >, < 40, 28 > .
Our group A7 acts irreducibility on C.
If Wi denote the set of all words in C of weight i, then
C =< W32 >=< W40 >,
so C is generated by its minimum-weight codewords. The full automorphism
group of C is Aut(C) ∼
= 235 :S8 with |Aut(C)| = 242 .32 .5.7, and we note that
Aut(C) ≥ Aut(D) and that Aut(D) is not a normal subgroup of Aut(C).
Furthermore C ⊥ is a [70, 64, 2]2 code and its weight distribution has been
determined. Since the blocks of D are of even size 40, we have that  meets
10
METHOD 2
68
l
32
|Wl |
35
40(1)
7
235 :S6
40(2)
21
235 :(S5 :2)
Aut(D)wl
235 :(A4 × 3):2
Table 8: Stabilizer of a word wl in Aut(D)
l
32
|Wl |
35
40
28
Aut(D)wl
235 :(S4 × S4 ):2
235 :(S6 × 2)
Table 9: Stabilizer of a word wl in Aut(C)
evenly every vector of C and hence  ∈ C ⊥ . If W̄i denote the set of all codewords
in C ⊥ of weight i, then |W̄2 | = 35,, |W̄3 | = 840, |W̄4 | = 14035 and
C ⊥ =< W̄3 >, dim(< W̄2 >= 35, dim(< W̄4 >= 63.
Let eij denote the 2-cycle (i, j) in S7 , where {i, j} = s(w̄2 ) is the support of a
codeword w̄2 ∈ W̄2 . Then eij (w̄2 ) = w̄2 , and < eij |{i, j} = s(w¯2 ), w̄2 ∈ W̄2 >=
235 .
Using MAGMA we can easily show that V = F270 is decomposable into indecomposable G-modules of dimension 40 and 30. We also have dim(Soc(V ) = 21
and
Soc(V ) =<  > ⊕C ⊕ C14 ,
where C is our 6-dimensional code and C14 is an irreducible code of dimension
14.
The structures the stabilizers Aut(D)wl and Aut(C)wl , where l ∈ {32, 40} are
listed in Table 8 and 9.
10.1.2
G = A7 , M = A6 and nX = 2A
Let G = A7 , M = A6 and nX = 2A. Then
b = [G : M ] = 7, v = |2A| = 105, k = |M ∩ 2A| = 45.
Also using the character table of A7 , we have χM = χ1 + χ2 = 1a + 6a and hence
χM (g) = 1+2 = 3 = λ, where g ∈ 2A. We produce a non-symmetric 1−(105, 45, 3)
design D. A7 acts primitively on the 7 blocks. Since CA7 (g) = D8 : 3 is not
maximal in A7 (sits in the maximal subgroup (A4 × 3):2 with index three), A7
acts imprimitively on the 105 points. The complement of D, D̃, is a 1−(105, 60, 4)
design.
The full automorphism group of D is
Aut(D) ∼
= S3 35 :S7 ∼
= S3 5 o S7 ,
10
METHOD 2
69
with |Aut(D)| = 242 .337 .5.7.
Construction using MAGMA shows that the binary code C of this design is a
[105, 7, 45]2 code. The weight distribution of C is
< 0, 1 >, < 45, 28 >, < 48, 35 >, < 57, 35 >, < 60, 28 >, < 105, 1 > .
We also have that Hull(C) is a [105, 6, 48] code and has the following weight
distribution:
< 0, 1 >, < 48, 35 >, < 60, 28 > .
Note that C = Hull(C)⊕ <  >, and that our group A7 acts irreducibility on
Hull(C). Also note that this result together with the result obtained in 5.1.2
imply that the 6-dimensional irreducible representation of A7 over GF (2) could
be represented by two non-isomorphic codes, namely [105, 6, 48]2 and [70, 6, 32]2
codes.
We also have
C =< W45 >=< W57 >,
so C is generated by its minimum-weight codewords. The full automorphism
group of C is Aut(C) = Aut(D) and its structure was given above in 5.2.1.
Using MAGMA we can easily show that V = F2105 is decomposable into indecomposable G-modules of dimension 1, 14, 20 and 70 (the first three are irreducible). We also have dim(Soc(V ) = 55 and that
Soc(V ) =<  > ⊕C14 ⊕ C14 ⊕ C20 ⊕ Hull(C),
where C = Hull(C)⊕ <  > is our 7-dimensional code and C14 and C20 are
irreducible codes of dimension 14 and 20 respectively.
10.1.3
G = A7 , M = S5 and nX = 2A: 1 − (105, 25, 5) Design
Let G = A7 , M = S5 and nX = 2A. Then
b = [G : M ] = 21, v = |2A| = 105, k = |M ∩ 2A| = 25.
Note that both conjugacy classes of involutions of S5 fuses to 2A. Also using
the character table of A7 , we have χM = χ1 + χ2 + χ5 = 1a + 6a + 14a and
hence χM (g) = 1 + 2 + 2 = 5 = λ, where g ∈ 2A. We produce a non-symmetric
1 − (105, 25, 5) design D. A7 acts primitively on the 21 blocks. Since CA7 (g) =
D8 :3 is not maximal in A7 (sits in the maximal subgroup (A4 × 3):2 with index
three), A7 acts imprimitively on the 105 points. The complement of D, D̃, is a
1 − (105, 80, 16) design.
10.1.4
G = A7 , M = P SL2 (7) and nX = 2A: 1 − (105, 21, 3) Design
Let G = A7 , M = P SL2 (7) and nX = 2A. Then
b = [G : M ] = 15, v = |2A| = 105, k = |M ∩ 2A| = 21.
Also using the character table of A7 , we have χM = χ1 + χ6 = 1a + 14b and hence
χM (g) = 1+2 = 3 = λ, where g ∈ 2A. We produce a non-symmetric 1−(105, 21, 3)
10
METHOD 2
70
design D. A7 acts primitively on the 15 blocks. Since CA7 (g) = D8 : 3 is not
maximal in A7 (sits in the maximal subgroup (A4 × 3):2 with index three), A7
acts imprimitively on the 105 points. The complement of D, D̃, is a 1−(105, 84, 12)
design.
10.1.5
G = A7 , M = P SL2 (7) and nX = 3B: 1 − (280, 56, 3) Design
Let G = A7 , M = P SL2 (7) and nX = 3B. Then
b = [G : M ] = 15, v = |3B| = 280, k = |M ∩ 2A| = 56.
Also using the character table of A7 , we have χM = χ1 + χ6 = 1a + 14b and hence
χM (g) = 1+2 = 3 = λ, where g ∈ 3B. We produce a non-symmetric 1−(280, 56, 3)
design D. A7 acts primitively on the 15 blocks. Since CA7 (g) = 3×3 ∈ Syl3 (A7 ) is
not maximal in A7 (sits in the maximal subgroups A6 and (A4 × 3):2 with indices
40 and 8 respectively), A7 acts imprimitively on the 280 points. The complement
of D, D̃, is a 1 − (280, 224, 12) design.
10.2
Design and codes from P SL2 (q)
The main aim of this section to develop a general approach to G = P SL2 (q), where
M is the maximal subgroup that is the stabilizer of a point in the natural action of
degree q +1 on the set Ω. This is fully discussed in Subsection 5.2.1. We start this
section by applying the results discussed for Method 1, particularly the Theorem
10.1, to all maximal subgroups and conjugacy classes of elements of P SL2 (11)
to construct 1- designs and their corresponding binary codes. These are itemized
bellow after Tables 5 and 6. The group P SL2 (11) has order 660 = 22 × 3 × 5 × 11,
it has four conjugacy classes of maximal subgroups, which are listed in the table
10. It has also eight conjugacy classes of elements which we list in Table 11.
No.
Max[1]
Max[2]
Max[3]
Max[4]
Order
55
60
60
12
Index
12
11
11
55
Structure
F55 = 11 : 5
A5
A5
D12
Table 10: Maximal subgroups of P SL2 (11)
Max[1]
5A: D = 1 − (132, 22, 2), b = 12; C = [132, 11, 22]2 , C ⊥ = [132, 121, 2]2 ;
Aut(D) = Aut(C) = 266 : S12 .
5B: As for 5A.
11A: D = 1 − (60, 5, 1), b = 12; C = [60, 12, 5]2 , C ⊥ = [60, 48, 2]2 ;
Aut(D) = Aut(C) = (S5 )12 : S12 .
11B: As for 11A.
10
METHOD 2
71
nX
2A
3A
5A
5B
6A
11A
11B
|nX|
55
110
132
132
110
60
60
CG (g)
D12
Z6
Z5
Z5
Z6
Z11
Z11
Maximal Centralizer
Yes
No
No
No
No
No
No
Table 11: Conjugacy classes of P SL2 (11)
Max[2]
2A : D = 1 − (55, 15, 3), b = 11; C = [55, 11, 15]2 , C ⊥ = [55, 44, 4]2 ;
Aut(D) = P SL2 (11), Aut(C) = P SL2 (11) : 2.
3A : D = 1 − (110, 20, 2), b = 11; C = [110, 10, 20]2 , C ⊥ = [110, 100, 2]2 ;
Aut(D) = Aut(C) = 255 : S11 .
5A : D = 1 − (132, 12, 1), b = 11; C = [132, 11, 12]2 , C ⊥ = [132, 121, 2]2 ;
Aut(D) = Aut(C) = (S12 )11 : S11 .
5B : As for 5A.
Max[3]
As for Max[2].
Max[4]
2A : D = 1 − (55, 7, 7), b = 55; C = [55, 35, 4]2 , C ⊥ = [55, 20, 10]2 ;
Aut(D) = Aut(C) = P SL2 (11) : 2.
3A : D = 1 − (110, 2, 1), b = 55; C = [110, 55, 2]2 , C ⊥ = [110, 55, 2]2 ;
Aut(D) = Aut(C) = 255 : S55 .
6A : As for 3A.
10.2.1
G = P SL2 (q) of degree q + 1, M = G1
Let G = P SL2 (q), let M be the stabilizer of a point in the natural action of
degree q + 1 on the set Ω. Let M = G1 . Then it is well known that G acts sharply
2-transitive on Ω and M = Fq : Fq∗ = Fq : Zq−1 , if q is even, and M = Fq : Z q−1 ,
2
if q is odd. Since G acts 2-transitively on Ω, we have χ = 1 + ψ where χ is
the permutation character of the action and ψ is an irreducible character of G
of degree q. Also since the action is sharply 2-transitive, only 1G fixes 3 distinct
elements of Ω. Hence for all 1G 6= g ∈ G we have λ = χ(g) ∈ {0, 1, 2}.
Proposition 10.2 For G = P SL2 (q), let M be the stabilizer of a point in the
natural action of degree q + 1 on the set Ω. Let M = G1 . Suppose g ∈ nX ⊆ G is
an element fixing exactly one point, and without loss of generality, assume g ∈ M .
Then the replication number for the associated design is r = λ = 1. We also have
10
METHOD 2
72
(i) If q is odd then |g G | = 12 (q 2 − 1), |M ∩ g G | = 12 (q − 1), and D is a 1( 21 (q 2 − 1), 12 (q − 1), 1) design with q + 1 blocks and
Aut(D) = S 12 (q−1) o Sq+1 = (S 12 (q−1) )q+1 : Sq+1 .
For all p, C = Cp (D) = [ 12 (q 2 − 1), q + 1, 12 (q − 1)]p , with Aut(C) = Aut(D).
(ii) If q is even then |g G | = (q 2 − 1), |M ∩ g G | = (q − 1), and D is a 1((q 2 − 1), (q − 1), 1) design with q + 1 blocks and
Aut(D) = S(q−1) o Sq+1 = (S(q−1) )q+1 : Sq+1 .
For all p, C = Cp (D) = [(q 2 − 1), q + 1, q − 1)]p , with Aut(C) = Aut(D).
Proof: Since χ(g) = 1, we deduce that ψ(g) = 0. We now use the character table
and conjugacy classes of P SL2 (q) (for example see [14]):
(i) For q odd, there are two types of conjugacy classes with ψ(g) = 0. In
both cases we have |CG (g)| = q and hence |nX| = |g G | = |P SL2 (q)|/q =
(q 2 − 1)/2. Since b = [G : M ] = q + 1 and
k=
1 × (q 2 − 1)/2
χ(g) × |nX|
=
= (q − 1)/2,
[G : M ]
q+1
the results follow from Remark 10.2.
(ii) For q even, P SL2 (q) = SL2 (q) and there is only one conjugacy
class with
1 0
ψ(g) = 0. A class representative is the matrix g =
with |CG (g)| =
1 1
q and hence |nX| = |g G | = |P SL2 (q)|/q = (q 2 −1). Since b = [G : M ] = q+1
and
1 × (q 2 − 1)
χ(g) × |nX|
=
= q − 1,
k=
[G : M ]
q+1
the results follow from Remark 10.2. If we have λ = r = 2 then a graph (possibly with multiple edges) can be
defined on b vertices, where b is the number of blocks, i.e. the index of M in G,
by stipulating that the vertices labelled by the blocks bi and bj are adjacent if bi
and bj meet. Then the incidence matrix for the design is an incidence matrix for
the graph.
In the case where the graph is an undirected graph without multiple edges the
following result from [8, Lemma] can be used.
Lemma 10.3 ([8]) Let Γ = (V, E) be a regular graph with |V | = N , |E| = e and
valency v. Let G be the 1-(e, v, 2) incidence design from an incidence matrix A
for Γ. Then Aut(Γ) = Aut(G).
Note: If the graph Γ is also connected, then it is an easy induction to show that
rankp (A) ≥ |V | − 1 for all p with obvious equality when p = 2. If in addition (as
happens for some classes of graphs, see [8, 25, 24]) the minimum weight is the
valency and the words of this weight are the scalar multiples of the rows of the
incidence matrix, then we also have Aut(Cp (G)) = Aut(G).
10
METHOD 2
73
Proposition 10.4 For G = P SL2 (q), let M be the stabilizer of a point in the
natural action of degree q + 1 on the set Ω. Let M = G1 . Suppose g ∈ nX ⊆ G
is an element fixing exactly two points, and without loss of generality, assume
g ∈ M = G1 and that g ∈ G2 . Then the replication number for the associated
design is r = λ = 2. We also have
(i) If g is an involution, so that q ≡ 1 (mod 4), the design D is a 1-( 21 q(q +
1), q, 2) design with q + 1 blocks and Aut(D) = Sq+1 . Furthermore C2 (D) =
[ 12 q(q + 1), q, q]2 , Cp (D) = [ 12 q(q + 1), q + 1, q]p if p is an odd prime, and
Aut(Cp (D)) = Aut(D) = Sq+1 for all p.
(ii) If g is not an involution, the design D is a 1-(q(q + 1), 2q, 2) design with
1
q + 1 blocks and Aut(D) = 2 2 q(q+1) : Sq+1 . Furthermore C2 (D) = [q(q +
1), q, 2q]2 , Cp (D) = [q(q+1), q+1, 2q]p if p is an odd prime, and Aut(Cp (D)) =
1
Aut(D) = 2 2 q(q+1) : Sq+1 for all p.
Proof: A block of the design constructed will be M ∩ g G . Notice that from
elementary considerations or using group characters we have that the only powers
of g that are conjugate to g in G are g and g −1 . Since M is transitive on Ω \ {1},
g M and (g −1 )M give 2q elements in M ∩ g G if o(g) 6= 2, and q if o(g) = 2.
These are all the elements in M ∩ g G since Mj is cyclic so if h1 , h2 ∈ Mj and
h1 = g1x , h2 = g2x for some x1 , x2 ∈ G, then h1 is a power of h2 , so they can only
be equal or inverses of one another.
(i) In this case by the above k = |M ∩ g G | = q and hence
|nX| =
q × (q + 1)
k × [G : M ]
=
.
χ(g)
2
So D is a 1-( 12 q(q + 1), q, 2) design with q + 1 blocks. An incidence matrix
of the design is an incidence matrix of a graph on q + 1 points labelled by
the rows of the matrix, with the vertices corresponding to rows ri and rj
being adjacent if there is a conjugate of g that fixes both i and j, giving
an edge [i, j]. Since G is 2-transitive, the graph we obtain is the complete
graph Kq+1 .
The automorphism group of the design is the same as that of the graph
(see [8]), which is Sq+1 . By [24], C2 (D) = [ 21 q(q + 1), q, q]2 and Cp (D) =
[ 12 q(q + 1), q + 1, q]p if p is an odd prime. Further, the words of the minimum
weight q are the scalar multiples of the rows of the incidence matrix, so
Aut(Cp (D)) = Aut(D) = Sq+1 for all p.
(ii) If g is not an involution, then k = |M ∩ g G | = 2q and hence
|nX| =
k × [G : M ]
2q × (q + 1)
=
= q(q + 1).
χ(g)
2
So D is a 1-(q(q + 1), 2q, 2) design with q + 1 blocks. In the same way we
define a graph from the rows of the incidence matrix, but in this case we
have the complete directed graph.
10
METHOD 2
74
1
The automorphism group of the graph and of the design is 2 2 q(q+1) : Sq+1 .
Similarly to the previous case, C2 (D) = [q(q + 1), q, 2q]2 and Cp (D) =
[q(q + 1), q + 1, 2q]p if p is an odd prime. Further, the words of the minimum
weight 2q are the scalar multiples of the rows of the incidence matrix, so
1
Aut(Cp (D)) = Aut(D) = 2 2 q(q+1) : Sq+1 for all p. We end this subsection by giving few examples of designs and codes constructed, using Propositions 10.2 and 10.4, from P SL2 (q) for q ∈ {16, 17, 19},
where M is the stabilizer of a point in the natural action of degree q + 1 and
g ∈ nX ⊆ G is an element fixing exactly one or two points.
Example 10.1 (P SL2 (16))
1. g is an involution having cycle type 11 28 , r = λ = 1: D is a 1 − (255, 15, 1)
design with 17 blocks. For all p, C = Cp (D) = [255, 17, 15]p , with Aut(C) =
Aut(D) = S15 o S17 = (S15 )17 : S17 .
2. g is an element of order 3 having cycle type 12 35 , r = λ = 2: D is a
1 − (272, 32, 2) design with 17 blocks. C2 (D) = [272, 16, 32]2 and Cp (D) =
[272, 17, 32]p for odd p. Also for all p we have Aut(Cp (D)) = Aut(D) =
2136 : S17 .
Example 10.2 (P SL2 (17))
Note that 17 ≡ 1 (mod 4).
1. g is an element of order 17 having cycle type 11 171 , r = λ = 1: D is a
1 − (144, 8, 1) design with 18 blocks. For all p, C = Cp (D) = [144, 18, 8]p ,
with Aut(C) = Aut(D) = S8 o S18 = (S8 )18 : S18 .
2. g is an involution having cycle type 12 28 , r = λ = 2: D is a 1 − (153, 17, 2)
design with 18 blocks. C2 (D) = [153, 17, 17]2 and Cp (D) = [153, 18, 17]p for
odd p. Also for all p we have Aut(Cp (D)) = Aut(D) = S18 .
3. g is an element of order 4 having cycle type 12 44 , r = λ = 2: D is a
1 − (306, 34, 2) design with 18 blocks. C2 (D) = [306, 17, 34]2 and Cp (D) =
[306, 18, 34]p for odd p. Also for all p we have Aut(Cp (D)) = Aut(D) =
2153 : S18 .
4. g is an element of order 8 having cycle type 12 82 , r = λ = 2: D is a
1 − (306, 34, 2) design with 18 blocks. C2 (D) = [306, 17, 34]2 and Cp (D) =
[306, 18, 34]p for odd p. Also for all p we have Aut(Cp (D)) = Aut(D) =
2153 : S18 .
Example 10.3 (P SL2 (19))
1. g is an element of order 19 having cycle type 11 191 , r = λ = 1: D is a
1 − (180, 9, 1) design with 20 blocks. For all p, C = Cp (D) = [180, 20, 9]p ,
with Aut(C) = Aut(D) = S9 o S20 = (S9 )20 : S20 .
2. g is an element of order 3 having cycle type 12 36 , r = λ = 2: D is a
1 − (380, 38, 2) design with 20 blocks. C2 (D) = [360, 19, 38]2 and Cp (D) =
[360, 20, 38]p for odd p. Also for all p we have Aut(Cp (D)) = Aut(D) =
2190 : S20 .
10
METHOD 2
10.3
75
Some 1-designs from the Janko group J1
The Jako group J1 of order 23 × 3 × 5 × 7 × 11 × 19 has seven conjugacy classes
of maximal subgroups, which were listed in the table 1. It has also 15 conjugacy
classes of elements some of which are listed in Table 12.
nX
2A
3A
|nX|
1463
5852
CG (g)
2 × A5
D6 × 5
Maximal Centralizer
Yes
No
Table 12: Some of the conjugacy classes of J1
We apply the Theorem 10.1 to the maximal subgroups and few conjugacy
classes of elements of J1 to construct several 1- designs.
10.3.1
G = J1 , M = P SL2 (11) and nX = 2A: 1 − (1463, 55, 10) Design
Let G = J1 , M = P SL2 (11) and nX = 2A. Then
b = [G : M ] = 266, v = |2A| = 1463, k = |M ∩ 2A| = 55.
Also using the character table of J1 , we have
χM = χ1 + χ2 + χ4 + χ6 = 1a + 56a + 56b + 76a + 77a
and hence χM (g) = 1 + 0 + 0 + 4 + 5 = 10 = λ, where g ∈ 2A. We produce a
non-symmetric 1−(1463, 55, 10) design D. Since CG (g) = 2×A5 is also a maximal
subgroup of J1 , J1 acts primitively on blocks and points. The complement of D,
D̃, is a 1 − (1463, 1408, 256) design.
10.3.2
G = J1 , M = 2 × A5 and nX = 2A: 1 − (1463, 31, 31) Design
Let G = J1 , M = 2 × A5 and nX = 2A. Then
b = [G : M ] = 1463, v = |2A| = 1463.
It is easy to see that M = 2 × A5 has three conjugacy classes of order 2, namely
x1 = z, x2 = α and x3 = zα, that fuse to 2A with corresponding centralizer
orders 120, 8 and 8. Now by using Corollary 6.3 we have
λ = χM (g) =
3
X
|CG (g)|
120 120 120
=
+
+
= 31,
|CM (xi )|
120
8
8
i=1
where g ∈ 2A. Alternatively we can use the character table of J1 to find that
χM = χ1 + χ2 + χ3 + 2χ4 + 2χ6 + χ9 + χ10 + χ11 + 2χ12 + 2χ15 ,
and
χM (g) = 1 + 0 + 0 + 8 + 10 + 0 + 0 + 0 + 10 + 2 = 31 = λ.
In this case clearly k = |M ∩ 2A| = λ = 31, and we produce a symmetric 1 −
(1463, 31, 31) design D. Obviously J1 acts primitively on blocks and points. The
complement of D, D̃, is a 1 − (1463, 1432, 1432) design.
REFERENCES
10.3.3
76
G = J1 , M = P SL2 (11) and nX = 3A: 1 − (5852, 110, 5) Design
Let G = J1 , M = P SL2 (11) and nX = 3A. Then
b = [G : M ] = 266, v = |3A| = 5852, k = |M ∩ 3A| = 110.
Also using the character table of J1 , we have
χM = χ1 + χ2 + χ4 + χ6 = 1a + 56a + 56b + 76a + 77a
and hence χM (g) = 1 + 4 + 1 − 1 = 5 = λ, where g ∈ 3A. We produce a nonsymmetric 1 − (5852, 110, 5) design D. Since CG (g) = D6 × 5 is not a maximal
subgroup of J1 , J1 acts primitively on 266 blocks but imprimitively on 5852 points.
The complement of D, D̃, is a 1 − (5852, 5742, 261) design.
10.3.4
G = J1 , M = P SL2 (11) and nX = 3A: 1 − (5852, 20, 5) Design
Let G = J1 , M = 2 × A5 and nX = 3A. Then
b = [G : M ] = 1463, v = |3A| = 5852, k = |M ∩ 3A| = 20.
It is easy to see that M = 2 × A5 has only one conjugacy class of elements of
order 3, which fuses to 3A, with the corresponding centralizer order 6. Now by
using Corollary 6.3 we have
λ = χM (g) =
30
|CG (g)|
=
= 5,
|CM (x)|
6
where g ∈ 3A. Alternatively we can use the character χM as in Subsection 10.3.2
to find that
χM (g) = 1 + 2 + 2 + 2 − 2 + 0 + 0 + 0 + 2 − 2 = 5 = λ,
where g ∈ 3A. We produce a non-symmetric 1 − (5852, 20, 5) design D. Since
CG (g) = D6 × 5 is not a maximal subgroup of J1 , J1 acts primitively on the
1463 blocks but imprimitively on the 5852 points. The complement of D, D̃, is a
1 − (5852, 5832, 1458) design.
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